azl397985856 commented 3 years ago

69. x 的平方根

入选理由

暂无

题目地址

https://leetcode-cn.com/problems/sqrtx

前置知识

二分法
题目描述
```
实现 int sqrt(int x) 函数。
```

计算并返回 x 的平方根，其中 x 是非负整数。

由于返回类型是整数，结果只保留整数的部分，小数部分将被舍去。

示例 1:

输入: 4 输出: 2 示例 2:

输入: 8 输出: 2 说明: 8 的平方根是 2.82842..., 由于返回类型是整数，小数部分将被舍去

james20141606 commented 3 years ago

Day 37: 69. Sqrt(x) (binary search, math, newton method)

Problem Link
- 69. Sqrt(x)
Ideas
- Binary search, from 1 to x, return left, my condition is if middle ** 2 <= x and if (middle + 1) ** 2 > x: return middle
- We could also try newton's method for this iteration problem.
- We could solve f(x) =0 using newton's method. For the function f(x) we could have the tangent line at $(\mathrm{x_k}, \mathrm{f}(\mathrm{xk}))$ which is $y=f\left(x{k}\right)+f^{\prime}\left(x{k}\right)\left(x{k+1}-x_{k}\right)$. Intersection with x axis is $(xk+1,0)$. It satisfies: $0=f\left(x{k}\right)+f^{\prime}\left(x{k}\right)\left(x{k+1}-x_{k}\right)$. If the $f^{\prime}(xk) \neq 0$, then we have $x{k+1}=x{k}-\frac{f\left(x{k}\right)}{f^{\prime}\left(x_{k}\right)}$.
- As k increases, $\mathrm{xk}$ will approximate $\mathrm{x}^{*}$, which is $\lim {k \rightarrow 0} x_{k}=x^{*}$
- For our problem, we need positive root of $f(x)=x^ 2-a=0 \quad(a>0)$. Using newton, $f(x)=2 \cdot x$, so $x^{k+1}=x{k}-\frac{f\left(x{k}\right)}{f^{\prime}\left(x{k}\right)}=x{k}-\frac{x{k}^{2}-a}{2 x{k}}=\frac{1}{2}\left(x{k}+\frac{a}{x{k}}\right)$
Complexity: hash table and bucket
- Time: O(logN)
- Space: O(1)
Code


class Solution:
    def mySqrt(self, x: int) -> int:
        if x == 0: return 0 
        left, right = 1, x 
        while left < right:
            middle = (left + right) // 2
            #print (middle)
            if middle ** 2 <= x:
                if (middle + 1) ** 2 > x:
                    return middle
                left = middle
            else:
                right = middle
        return left
#1, x   , left = middle, right = middle, return left

class Solution:
    def mySqrt(self, x: int) -> int:
        ans, l, r = 0, 0, x
        while l <= r:
            mid = (l + r) // 2
            if mid ** 2 > x:
                r = mid - 1
            if mid ** 2 <= x:
                ans = mid
                l = mid + 1
        return int(ans)

class Solution:
    def mySqrt(self, x: int) -> int:    
        r = x + 1  # avoid dividing 0
        while r*r > x:
            r = int(r - (r * r - x) / (2 * r))  # newton's method
        return r

other resources:
- https://leetcode-cn.com/problems/sort-an-array/solution/pai-xu-shu-zu-by-leetcode-solution/

shixinlovey1314 commented 3 years ago

Title:69. Sqrt(x)

Question Reference LeetCode

Solution

Binary search,

Code

class Solution {
public:
    int mySqrt(int x) {

        if (x < 2)
            return x;

        int lo = 2, hi = x / 2;

        while (lo <= hi) {
            long mid = lo + (hi - lo) / 2;
            long num = mid * mid;
            if (num == x)
                return mid;
            else if (num > x)
                hi = mid - 1;
            else
                lo = mid + 1;
        }

        return lo - 1;
    }
};

Complexity

Time Complexity and Explanation

O(logx)

Space Complexity and Explanation

O(1)

zhy3213 commented 3 years ago

思路

二分搜索

代码

    def mySqrt(self, x: int) -> int:
        if x==0 or x==1:
            return x
        l, r = 0, x
        res=(l+r)//2
        while not(res*res<=x and (res+1)*(res+1)>x):
            if res*res>x:
                r=res
            else:
                l=res
            res=(l+r)//2
        return res

一次摸奖99%

KennethAlgol commented 3 years ago

思路

二分法

class Solution {
    public int mySqrt(int x) {
        int left = 0;
        int right = x;

        while(left < right){
            int mid = left + (right - left) / 2;  //向下取整
            if(mid * mid >= x){
                right = mid;
            }else{
                left = mid + 1;
            }
        }

        return (long)left * left == x ? left : left -1 

    }
}

itsjacob commented 3 years ago

Intuition

using binary search template for right-bound search

Implementation

class Solution
{
public:
  int mySqrt(int x)
  {
    if (x == 0) return 0;
    int left{ 1 };
    int right{ x };
    while (left <= right) {
      int mid = left + (right - left) / 2;
      int x1 = x / mid;
      if (x1 == mid) {
        left = mid + 1;
      } else if (x1 > mid) {
        left = mid + 1;
      } else if (x1 < mid) {
        right = mid - 1;
      }
    }
    return right;
  }
};

Complexity

Time complexity: O(logx)
Space complexity: O(1)

newbeenoob commented 3 years ago

思路

搜索右侧插入位置的二分模板（右侧边界，即搜索的是最后一个满足 v ≤ ∟ x / v 的位置），+1的操作主要是为了避免发生死循环（一个补丁），因为 >> 等效于向下取整，例如 l = 0 , r = 1 时候 0 + 1 / 2 = 0 , l 仍是0，就会发生死循环。

代码：JavaScript

var mySqrt = function(x) {
    let l = 0;
    let r = Math.ceil(x / 2); // 搜索区间的定界 这里定义为 x / 2 可以缩小一半初始搜索区间 不影响正确答案
    let mid;
    while(l < r) {
        mid = ((r - l) >> 1) + l + 1; // + 1 补丁 避免死循环
        if (mid <= Math.floor(x / mid)) {
            l = mid;
        } else {
            r = mid - 1;
        }
    }

    return l;
};

复杂度分析

时间复杂度： O(logn)
额外空间复杂度： O(1)

class Solution(object):
    def mySqrt(self, x):
        """
        :type x: int
        :rtype: int
        """
        low, up = 0, x
        while low**2 <= x:
            mid = (low+up) // 2
            if mid**2<=x and (mid+1)**2>x:
                return mid
            elif mid**2>x:
                up = mid - 1
            else:
                low = mid + 1

复杂度

时间复杂度：O(logx)
空间复杂度：O(1)

thinkfurther commented 3 years ago

思路

使用二分法

代码

class Solution:
    def mySqrt(self, x: int) -> int:
        leftValue = 0
        rightValue = x

        while(leftValue <= rightValue):
            middleValue = (leftValue + rightValue) // 2
            print(middleValue)

            if(middleValue * middleValue > x):
                rightValue = middleValue - 1
            elif(middleValue * middleValue < x):
                leftValue = middleValue + 1
            else:
                return middleValue

        middleValue = (leftValue + rightValue) // 2
        return middleValue

复杂度

时间复杂度：O(log N)

空间复杂度：O(1)

xbhog commented 3 years ago

69. Sqrt(x)

思路：

采用二分模板,找到中间值，如果mid相乘减去原来的数小于一个精度值，那么我们就认为是开平方的结果。

Java代码段：

class Solution {
    public int mySqrt(int x) {
        double n = (double)x;
        double l = 0,r = n;

        while(l <= r){
            double mid = (l+r)/2;
            if(Math.abs(mid*mid-n) < 1e-6) return (int)mid;

            if(mid*mid-n > 1e-6) r = mid-1e-6;
            else l = mid+1e-6;
        }
        return (int)l;
    }
}

复杂度分析：

时间复杂度：O(logn)

空间复杂度：O(1)

chenming-cao commented 3 years ago

解题思路

二分法。寻找最右边的满足条件的值。

代码

class Solution {
    public int mySqrt(int x) {
        int l = 0, r = x;
        int mid = 0;
        int ans = -1;
        while (l <= r) {
            mid = l + (r - l) / 2;
            if ((long) mid * mid <= x) { // convert to long to avoid overflow
                ans = mid;
                l = mid + 1;
            }
            else {
                r = mid - 1;
            }
        }
        return ans;
    }
}

复杂度分析

时间复杂度：O(logx)
空间复杂度：O(1)

weichuliao commented 3 years ago

Code

class Solution:
    def mySqrt(self, x: int) -> int:
        ans, l, r = 0, 0, x
        while l <= r:
            mid = (l + r) // 2
            if mid ** 2 > x:
                r = mid - 1
            if mid ** 2 <= x:
                ans = mid
                l = mid + 1
        return int(ans)

Complexity Analysis

Time:
Space:

Zhang6260 commented 3 years ago

JAVA版本

思路：核心使用二分，但是需要有几点注意，如果使用乘法作为二分条件时候，需要考虑是否越界的问题(也可以使用除法，x/mid与mid进行比较).

class Solution {
   public int mySqrt(int x) {
       if(x==0)return 0;

       int left = 1,right = 46340;
       while(left<=right){
           int mid = (right -left)/2 + left;
           if(mid*mid==x){
               return mid;
           }else if(mid*mid<x){
               left = mid+1;

           }else{
               right = mid -1;
           }
       }
       return right;
   }
}

时间复杂度：O(logn)

空间复杂度：O(1）

Francis-xsc commented 3 years ago

思路

二分法，注意爆int的问题

代码


class Solution {
public:
    int mySqrt(int x) {
        int l=0,r=x;
        int ans=-1;
        while(l<=r)
        {
            int mid=(l+r)>>1;
            long long int p=(long long)mid*mid;
            if(p==x)
                return mid;
            else if(p<x)
            {
                ans=mid;
                l=mid+1;
            }
            else if(p>x)
                r=mid-1;
        }
        return ans;
    }
};

复杂度分析

时间复杂度：O(logN)
空间复杂度：O(1)

biancaone commented 3 years ago

class Solution:
    def mySqrt(self, x: int) -> int:
        left, right = 0, x

        # find the last ele, that ele * ele <= x

        while left + 1 < right:
            mid = (left + right) // 2
            if mid * mid == x:
                return mid
            elif mid * mid > x:
                right = mid
            else:
                left = mid

        if right * right <= x:
            return right
        return left

septasset commented 3 years ago

Day 37

思考

关键点

代码(Python)

class Solution:
    def mySqrt(self, x: int) -> int:
        # 闭合区间 [l，r]
        l = 0
        r = x
        while l<=r:
            mi = (l+r) // 2
            if mi*mi == x: return mi
            elif mi*mi > x:
                r = mi
            else:
                if (mi+1)*(mi+1) > x: 
                    return mi
                else:
                    l = mi + 1
        return -1

复杂度分析

时间：O(logx)
空间：O(1)

TimmmYang commented 3 years ago

思路

二分

代码

class Solution:
    def mySqrt(self, x: int) -> int:
        l, r = 0, x
        res = 0
        while l <= r:
            mid = (l + r) // 2
            if mid ** 2 <= x:
                res = mid
                l = mid + 1
            else:
                r = mid - 1
        return res

复杂度

时间：O(log x) 空间：O(1)

siyuelee commented 3 years ago

二分：寻找最右边的满足n^2 <= x 的值

class Solution(object):
    def mySqrt(self, x):
        """
        :type x: int
        :rtype: int
        """
        res, l, r = 0, 0 ,x
        while l <= r:
            mid = (l + r) // 2
            if mid ** 2 == x:
                res = mid
                l = mid + 1
            if mid ** 2 < x:
                res = mid
                l = mid + 1
            if mid ** 2 > x:
                r = mid - 1
        return res

T: logx S: 1

JK1452470209 commented 3 years ago

思路

二分搜索

代码

    class Solution {
        public int mySqrt(int x) {
            int left = 0,right = x;
            if(x == 0 || x == 1){
                return x;
            }
            while(left + 1 < right){
                int mid = left + (right - left)/2;
                if(mid >= x/mid){
                    if(mid == x/mid){
                        return mid;
                    }
                    right = mid;
                }else{
                    left = mid;
                }
            }
            return left;
        }
    }

复杂度

时间复杂度：O(logN)

空间复杂度：O(1)

wangyifan2018 commented 3 years ago

class Solution:
    def mySqrt(self, x: int) -> int:
        left, right, ans = 0, x, 0
        while left <= right:
            mid = (left + right) // 2
            if (mid * mid) <= x:
                ans = mid
                left = mid + 1
            else:
                right = mid - 1
        return ans

qixuan-code commented 3 years ago

二分搜索

class Solution:
    def mySqrt(self, x: int) -> int:
        left, right = 0, x-1
        res = 0
        if x == 0: return 0
        if x == 1: return 1
        while left <= right:
            mid = left + (right-left)//2
            if mid**2 < x: #it means that sqrt(x)>mid,start from mid+1,right
                res = mid
                left = mid + 1
            elif mid**2 > x:
                right = mid -1
            elif mid**2 == x:
                return mid

        return res

ghost commented 3 years ago

题目

Sqrt(x)

思路

Binary Search

代码


class Solution:
    def mySqrt(self, x: int) -> int:

        if x <= 1: return x

        left, right = 0,x  
        while(left < right):
            mid = (left + right)//2
            if mid * mid <= x: 
                left =  mid + 1
            else:
                right = mid

        return left-1

复杂度

Space: O(1) Time: O(logx)

Yufanzh commented 3 years ago

Intuition

convert this problem to 'find the maximum value in [0...x-1] list where val^2 <= x' Using the "find_right_boundary" frame and pay attention to check boundary condition.

Algorithm in python3

class Solution:
    def mySqrt(self, x: int) -> int:
        #binary search approach, finding the rightest answer where i^2 <= x
        left = 0
        right = x-1
        while left <= right:
            mid = int(left + (right - left)/2)
            if mid * mid > x:
                right = mid - 1
            elif mid * mid == x:
                left = mid + 1
            elif mid * mid < x:
                left = mid + 1

        # boundary condition
        if x <= 1:
            return x
        return right

Complexity Analysis:

Time complexity: O(logN)
Space complexity: O(1)

ymkymk commented 3 years ago

思路

直接调用现成的方法

代码

``

class Solution {

    /**
     * 二分查找
     * @param x
     * @return
     */
    public int mySqrt(int x) {
        int l = 0, r = x/2 + 1, ans = -1;
        while (l <= r) {
            int mid = l + (r - l) / 2;
            if ((long) mid * mid <= x) {
                ans = mid;
                l = mid + 1;
            } else {
                r = mid - 1;
            }
        }
        return ans;
    }
}

复杂度分析

时间复杂度：O(logn)

空间复杂度：O(1)

zol013 commented 3 years ago

直接二分查找 Python 3 code

class Solution:
    def mySqrt(self, x: int) -> int:
        if x < 2:
            return x

        l, r = 2, x // 2
        while l <= r:
            mid = (l + r) // 2
            if mid * mid == x:
                return mid
            elif mid * mid < x:
                l = mid + 1
            else:
                r = mid - 1
        return l - 1

TC: O(logn) SC: O(1)

watermelonDrip commented 3 years ago

class Solution:
    def mySqrt(self, x: int) -> int:
        if x <= 1:
            return x
        l = 0
        r = x//2
        while l<=r:
            mid = (l+r)//2
            if mid**2 == x:
                return mid
            elif mid**2 >x:
                r = mid -1
            elif mid**2 < x:
                l = mid + 1
        return r

L-SUI commented 3 years ago

var mySqrt = function (x){ let l = 0; let r = x while( l < r ){ let mid = l+r+1 >> 1 if(mid <= x / mid){ l = mid }else{ r = mid - 1 } } return l }`

taojin1992 commented 3 years ago

Plan & Evaluate:

Plan:
binary search
time: O(logx)
space: O(1)

3

0 3
mid = 1
l = 2
mid = 2
r = 1

Tests: 
2147395599 -> 46339 
mid * mid overflow

8 
l 3
r 2
m 2

Code:

class Solution {
    public int mySqrt(int x) {
        int l = 1, r = x; // l=0 won't work for 0 case
        while (l <= r) {
            int mid = l + (r - l) / 2;
            if (x / mid == mid) {
                return mid;
            } else if (mid < x / mid) {
                l = mid + 1;
            } else {
                r = mid - 1;
            }
        }
        return r;
    }
}

muimi commented 3 years ago

思路

二分法

代码

class Solution {
  public int mySqrt(int x) {
    if (x == 0 || x == 1) return x;
    int left = 1, right = x;
    while (left < right) {
      // 特殊情况: x = Math.pow(2, 31); 如果left = 0, middle为负值
      int middle = left + (right - left + 1) / 2;
      // if判断跟向上取整还是向下取整没有关系
      if ((long)middle * middle <= x) {
        // 由于不判断等号，此时的middle有可能为解
        left = middle;
      } else {
        right = middle - 1;
      }
    }
    return left;
  }
}

复杂度

时间复杂度：O(logN)
空间复杂度：O(1)

LareinaWei commented 3 years ago

Thinking

Binary Search

Code

class Solution:
    def mySqrt(self, x: int) -> int:
        if x == 0:
            return 0
        l = 1
        r = x
        ans = 0
        while l <= r:
            mid = l + (r - l) // 2
            if mid ** 2 > x:
                r = mid - 1
            if mid ** 2 <= x:
                ans = mid
                l = mid + 1

        return ans

Complexity

Time complexity: O(logn). Space complexity: O(1)

zhangyalei1026 commented 3 years ago

思路

Binary search

代码

class Solution:
    def mySqrt(self, x: int) -> int:
        left, right = 0, x
        while left <= right:
            mid = (left + right) // 2
            if mid * mid > x:
                right = mid -1
            elif mid * mid < x:
                left = mid + 1
            else:
                return mid
        return right

复杂度分析

时间复杂度：O(logn) 空间复杂度：O(1)

ginnydyy commented 3 years ago

Problem

https://leetcode.com/problems/sqrtx/

Notes

binary search
look for the biggest integer k where k * k <= x
start is 2, end is x/2
since x could be the Integer.MAX_VALUE, x/2 x/2 could be int overflow, so have to use long for the result of mid mid
binary search template while (start + 1 < end) and int mid = start + (end - start)/2
when exit the while loop, start is before end, since looking for the biggest integer, so check end first

Solution

class Solution {
    public int mySqrt(int x) {
        if(x == 1){
            return 1;
        }

        int start = 2;
        int end = x / 2; // sqt(x) < x/2

        while(start + 1 < end){
            int mid = start + (end - start) / 2;
            long num = (long)mid * mid; // x <= 2^31 - 1, so x/2 * x/2 could be int overflow
            if(num > x){
                end = mid;
            }else if(num < x){
                start = mid;
            }else{
                return mid;
            }
        }

        // looking for the biggest integer k where k * k < x, so check end before start
        long num = (long)end * end;
        if(num <= x){
            return end;
        }

        return start;
    }
}

Complexity

Time: O(logx).
Space: O(1).

ai2095 commented 3 years ago

LC69. Sqrt(x)

https://leetcode.com/problems/sqrtx/

Topics

sort

思路

binary search

代码 Python

class Solution:
    def mySqrt(self, x: int) -> int:
        if x <2:
            return x
        left, right = 0, x
        while left <= right:
            mid = left + (right -left)//2
            if mid** 2 > x:
                right = mid -1
            elif mid** 2 < x:
                left = mid + 1
            else:
                return mid
        return right

复杂度分析

时间复杂度： O(logn)
空间复杂度：O(1)

carsonlin9996 commented 3 years ago

思路

二分法，根号x 必定是 0-x之间，二分法找最近(round down)


class Solution {
    public int mySqrt(int x) {

        long low = 0;
        long high = x;

        while (low + 1 < high) {
            long mid = low + (high - low) / 2;

            if (mid * mid == x) {
                return (int)mid;
            }

            if (mid * mid > x) {
                high = mid;
            } else {
                low = mid;
            }

        }
        if (high * high <= x) {
            return (int)high;
        }
        return (int)low;
    }
}

ZhuMengCheng commented 3 years ago

思路:二分法


var mySqrt = function (x) {
    let left = 0
    let right = x
    while (left <= right) {
        let mid = left + ((right - left) >> 1)
        if (mid * mid <= x) {
            left = mid + 1
        } else {
            right = mid - 1
        }
    }
    return right
};

时间复杂度：O(logn) 空间复杂度：O(1)

Bochengwan commented 3 years ago

思路

利用二分法。

代码

class Solution:
    def mySqrt(self, x: int) -> int:

        start,end = 0,x
        while start<=end:
            mid = (end+start)//2                         
            if mid**2<=x:
                start = mid+1
            else:
                end=mid-1

        return end

复杂度分析

时间复杂度：O(LOGN)，其中 N 为数组长度。
空间复杂度：O(1)

JiangyanLiNEU commented 3 years ago

Binary Search

square root can never be greater than half of x, so we can shrink the range
Runtime = O(logx)

Space = O(1)

JS Code

var mySqrt = function(x) {
let low = 0;
let high = Math.floor(x/2)+1;
let middle = 0;
while (low**2 < x && high**2 > x && (high-low)>1){
    middle = Math.floor((low+high)/2);
    if (middle**2 > x){
        high = middle-1;
    }else if (middle**2 === x){
        return middle;
    }else{
        low = middle;
    };
};
return high**2 <= x ? high : low;
};

15691894985 commented 3 years ago

【day37】69. x 的平方根

https://leetcode-cn.com/problems/sqrtx

二分法：注意一点：题中由于向下是取整，因此找的是满足 x^2<=8 的最大值，返回的是right

    class Solution:
        def mySqrt(self, x: int) -> int:
    #     二分法
        left = 0
        right = x
        while left <=right:
            mid = left +(right - left)//2
            if mid**2 > x:
                right = mid -1
            elif mid**2 < x:
                left = mid+1
            else:
                return mid
        return right

复杂度：

时间复杂度：O(log n)

空间复杂度：O(1)

V-Enzo commented 3 years ago

思路

直接从1遍历到x/2

0或1时需要进行额外操作。

class Solution {
public:
int mySqrt(int x) {
    if(x==0) return 0;
    if(x==1) return 1;
    for(long int i=1;i <=x/2; i++){
        if(i*i==x){
            return i;
        }
        if(i*i<x && (i+1)*(i+1)>x){
            return i;
        }
    }
    return 0;
}
};

还有二分法

class Solution {
public:
int mySqrt(int x) {
    if(x==1) return 1;
    int min = 0;
    int max = x;
    while(max - min > 1){
        int m = (max+min)/2;
        if(x/m<m){
            max = m;
        }else{
            min = m;
        }
    }
    return min;
}
};

Complexity:

Time:O(n) Space:O(1)

cy-sues commented 3 years ago

class Solution:
    def mySqrt(self, x: int) -> int:

        start,end = 0,x
        while start<=end:
            mid = (end+start)//2                         
            if mid**2<=x:
                start = mid+1
            else:
                end=mid-1

        return end

user1689 commented 3 years ago

题目

https://leetcode-cn.com/problems/sqrtx/

思路

二分区间

python

class Solution:
    def mySqrt(self, x: int) -> int:

        # time logn
        # space 1
        # 二分区间
        # 
        if x == 0: return 0
        left, right = 1, x
        while left < right:
            mid = left + (right - left + 1) // 2
            if mid * mid > x:
                right = mid - 1
            else:
                left = mid 
        return left

复杂度分析

time logn
space 1

思想

二分法

代码

  public int mySqrt(int x) {
        int l = 0, r = x, ans = -1;
        while (l <= r) {
            int mid = l + (r - l) / 2;
            if ((long) mid * mid <= x) {
                ans = mid;
                l = mid + 1;
            } else {
                r = mid - 1;
            }
        }
        return ans;
    }

mixtureve commented 3 years ago

思路

二分法

代码

语言支持：Java

Java Code:


public class Solution {

    public int mySqrt(int x) {

        if (x == 0) {
            return 0;
        }
        if (x == 1) {
            return 1;
        }

        int left = 1;
        int right = x / 2;
        while (left < right) {
            int mid = left + (right - left + 1) / 2;
            // 注意：这里为了避免乘法溢出，改用除法
            if (mid > x / mid) {
                right = mid - 1;
            } else {
                left = mid;
            }
        }
        return left;
    }
}

复杂度分析

令 n 为数组长度。

时间复杂度：$O(logn)$
空间复杂度：$O(1)$

kennyxcao commented 3 years ago

69. Sqrt(x)

Intuition

Binary Search

Code

/**
 * @param {number} x
 * @return {number}
 */
const mySqrt = function(x) {
  if (x < 2) return x;
  let start = 1;
  let end = ~~(x / 2);
  while (start <= end) {
    const mid = start + ~~((end - start) / 2);
    if (mid * mid <= x) {
      start = mid + 1;
    } else {
      end = mid - 1;
    }
  }
  return end;
};

Complexity Analysis

Time: O(logn)
Space: O(1)

Richard-LYF commented 3 years ago

class Solution: def mySqrt(self, x: int) -> int: if x == 0: return 0 ans = int(math.exp(0.5 * math.log(x))) return ans + 1 if (ans + 1) ** 2 <= x else ans

ChenJingjing85 commented 3 years ago

思路

二分搜索找到最右边的<=target的值

代码

class Solution {
    public int mySqrt(int x) {
        if(x == 1){
            return 1;
        }
        int l = 1;
        int r = x/2;
        while(l<=r){
            int mid = l+(r-l)/2;
            if((long) mid * mid == x){
                return mid;
            }
            if((long) mid * mid < x){
                l = mid+1;
            }else{
                r = mid-1;
            }
        }
        return r;

    }
}

复杂度分析

时间复杂度 O（logN）
空间复杂度 O（1）

xjlgod commented 3 years ago

class Solution {
    public int mySqrt(int x) {
        int res = 0;
        int left = 0, right = 46340;
        while (left <= right) {
            int mid = (left + right) / 2;
            long temp = mid * mid;
            if (temp == x) {
                return mid;
            } else if (temp > x) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        return right;
    }
}

asterqian commented 3 years ago

思路

二分查找注意都为闭区间[left, right]

代码

class Solution {
public:
    int mySqrt(int x) {
        int left = 0;
        int right = x;
        int ans = -1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if ((long long)mid * mid <= x) {
                ans = mid;
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return ans;
    }
};

复杂度分析

时间复杂度: O(logn)

空间复杂度: O(1)

mokrs commented 3 years ago

//二分查找，查找区间[0, x]
int mySqrt(int x) {
    int left = 0, right = x, res = 0;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        //由于结果向下取整，因此等于时的情况与小于时的合并
        if ((long long)mid * mid <= x) {
            res = mid;
            left = mid + 1;
        }
        else {
            right = mid - 1;
        }
    }

    return res;
}

复杂度：

时间：O(logn)
空间：O(1)

mannnn6 commented 3 years ago

代码

class Solution {
    public int mySqrt(int x) {
        if (x == 0) {
            return 0;
        }
        int ans = (int) Math.exp(0.5 * Math.log(x));
        return (long) (ans + 1) * (ans + 1) <= x ? ans + 1 : ans;
    }
}

复杂度

时间复杂度：O(logn) 空间复杂度：O(1)

ZJP1483469269 commented 3 years ago

思路

二分

代码


class Solution {
    public int mySqrt(int x) {
        if(x==0) return 0;
        if(x==1) return 1;
        int l =0,r=x;
        int mid;
        while(l<=r){
            mid = l + (r - l)/2;
            if(x/mid==mid) return mid;
            if(x/mid>mid) l = mid + 1;
            if(x/mid<mid) r = mid - 1;
        }
        return l-1;
    }
}

复杂度分析

时间复杂度：O（logn）空间复杂度：O（1）

leetcode-pp / 91alg-5-daily-check

【Day 37 】2021-10-16 - 69. x 的平方根 #54

69. x 的平方根

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题目描述

Day 37: 69. Sqrt(x) (binary search, math, newton method)

Title:69. Sqrt(x)

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