【Day 37 】2022-01-17 - 69. x 的平方根

[azl397985856]

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

[ZZRebas]

思路

二分，找所有满足 ans ^ 2 <= 8 的最大值，也就是找最右边的满足条件的值。

代码(Python)

def fun(x):
    l=0
    r=x
    while(l<=r):
        mid=(l+r)//2
        if mid**2 >x:
            r=mid-1
        elif mid**2 <=x:
            l=mid+1
            ans=mid
    return int(ans)
x=24
print(fun(x))

复杂度分析

• 时间复杂度：O(log(x))。
• 空间复杂度：

[falconruo]

思路:

二分查找

复杂度分析:

• 时间复杂度: O(logx)
• 空间复杂度: O(1)

代码(C++):

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

        while (l <= r) {
            long mid = l + (r - l)/2;

            if ((mid * mid <= x) && ((mid + 1)*(mid + 1) > x))
                return int(mid);
            else if (mid * mid > x)
                r = mid - 1;
            else
                l = mid + 1;
        }

        return l;
    }
};

[uniqlell]

思路：使用二分法

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

[luhnn120]

思路

二分法求解

代码

/**
 * @param {number} x
 * @return {number}
 */
var mySqrt = function(x) {
  if(x <= 1){
    return x
  }
  let i = Math.ceil(x/2)
  while(i > 0){
    if(i*i <= x) {
      return i
    }
    let j = Math.ceil(i / 2);
    if(j*j > x){
      i = j
    }else {
      i--
    }
  }
};

复杂度

空间复杂度 O(1) 时间复杂度 O(logn)

[zhiyuanpeng]

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

[jiaqiliu37]

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:
                ans = mid
                l = mid + 1
            else:
                r = mid - 1

        return ans

Time complexity O(log x) Space complexity O(1)

[zol013]

class Solution:
    def mySqrt(self, x: int) -> int:
        l, r = 0, x

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

        return r

[alongchong]

class Solution {
    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;
    }
}

[Myleswork]

思路

牛顿迭代法

代码

class Solution {
public:
    int mySqrt(int x) {
        //牛顿迭代
        if(x == 0){
            return 0;
        }
        double x0 = x;
        double c = x;
        while(true){
            //牛顿迭代法迭代
            double xi = 0.5*(c/x0+x0);
            if(x0-xi<1e-7){
                break;
            }
            x0 = xi;
        }
        return x0;

    }
};

复杂度分析

时间复杂度：O(logn)

空间复杂度：O(1)

[JudyZhou95]

思路

二分法，找到第一个平方后大于target的数，答案是这个数-1。

代码

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

    l, r = 0, x

    while l < r:
        mid = (l+r)//2            
        if mid*mid > x:
            r = mid
        else:
            l = mid + 1

    return r-1

复杂度

TC: O(logx) SC: O(1)

[Richard-LYF]

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

    l = 0 
    r = x 

    while r >= l:

        mid = l + (r-l)//2

        if mid*mid>x:
            r = mid -1 
        elif mid*mid<x:
            l = mid + 1
        else:
            return mid

    return r

[kite-fly6618]

思路

二分

代码

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
};

复杂度

时间复杂度:O(logn)
空间复杂度:O(1)

[ivangin]

code:

public int mySqrt2(int x) {
        long res = x;
        while (res * res > x) {
            res = (res + x / res) / 2;
        }
        return (int) res;
    }

[Neal0408]

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

[guoling0019]

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;
    }
}

复杂度分析

令 n 为数组长度。

• 时间复杂度：$O(1)$
• 空间复杂度：$O(1)$

[z1ggy-o]

思路

数学题，用的牛顿迭代。

代码

CPP

class Solution {
public:
    int mySqrt(int x) {
        long long c = x;
        while (c * c > x)
            c = (c + x / c) / 2;
        return (int)c;
    }
};

复杂度分析

• 时间：O(logx)
• 空间：O(1)

[LAGRANGIST]

Day 37-69. x 的平方根

Problem

69. Sqrt(x)

难度简单

给你一个非负整数 x ，计算并返回 x 的 算术平方根 。

由于返回类型是整数，结果只保留 整数部分 ，小数部分将被 舍去 。

注意：不允许使用任何内置指数函数和算符，例如 pow(x, 0.5) 或者 x ** 0.5 。

示例 1：

输入：x = 4
输出：2

示例 2：

输入：x = 8
输出：2
解释：8 的算术平方根是 2.82842..., 由于返回类型是整数，小数部分将被舍去。

提示：

• 0 <= x <= 231 - 1

思路

• 二分法找到满足 $mid^2 < x$的最大 $mid$

代码

class Solution
{
public:
    int num;

    int mySqrt(int x)
    {
        if(x <0) throw "Negative number doesn't have square root.";
        long l = 0, r = x;
        long mid = l + (r - l) / 4;
        long mid_square = 0;
        while (l <= r)
        {
            mid_square = mid * mid;
            if (mid_square == x)
                return mid;
            else if (mid_square < x)
            {
                if ((mid + 1) * (mid + 1) > x)
                    return mid;
                else
                    l = mid + 1;
            }
            else if (mid_square > x)
            {
                r = mid - 1;
            }
            mid = l + (r - l) / 4;
        }
        return -1;//Not found.
    }
};

复杂度

• Time complexity：$O(n)$

• Space complexity: $O(1)$

[lilililisa1998]

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

[Okkband]

class Solution {
public:
    int mySqrt(int x) {
        if (x ==0) return 0;
        if (x < 4) return 1;
        int left = 1;
        int right = x / 2;
        while(left <= right){
            int mid = left + (right - left) / 2;
            if ((long long)mid * mid < x){
                left = mid+1;
            } else if ((long long)mid * mid > x) {
                right = mid -1;
            } else {
                return mid;
            }
        }
        return right;
    }
};

[biscuit279]

二分法

class Solution(object): def mySqrt(self, x): """ :type x: int :rtype: int """ l = 1 r = x ans = 0 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)

时间复杂度：O(logn) 空间复杂度：O(1)

[now915]

var mySqrt = function(x) {
    let l = 0, r = x, res = -1
  while(l <= r) {
    let mid = parseInt((l + r) / 2)
    if (mid * mid <= x) {
      res = mid
      l = mid + 1
    } else {
      r = mid - 1
    }
  } 
  return res
};

• 时间复杂度 O(logn)
• 空间复杂度 O(1)

[last-Battle]

代码

• 语言支持：C++

C++ Code:

class Solution {
public:
    int mySqrt(int x) {
        long start = 0, end = x;
        long mid = 0;
        while (start <= end) {
            mid = start + (end - start) / 2;
            if (mid * mid == x) {
                return mid;
            } else if (mid * mid > x) {
                end = mid - 1;
            } else if (mid * mid < x) {
                start = mid + 1;
            }
        }

        if (end > 0 && end * end - x < start * start - x) {
            return end;
        }

        return start;
    }
};

复杂度分析

• 时间复杂度：$O(logn)$
• 空间复杂度：$O(1)$

[ray-hr]

代码

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

[Aobasyp]

思路 二分法

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

时间复杂度：O(log_x) 空间复杂度：O(1)

[demo410]

• 牛顿法

  int x0 = s / 2;     
  if ( x0 != 0 ){
      int x1 = ( x0 + s / x0 ) / 2;   
  
      while ( x1 < x0 ){
          x0 = x1;
          x1 = ( x0 + s / x0 ) / 2;
      }
  
      return x0;
  }else{
      return s;
  }

[stackvoid]

思路

二分法

算法

class Solution {
    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;
    }
}

算法分析

O(logN) O（1）

[kbfx1234]

69. Sqrt(x)

// 1-17 cpp
class Solution {
public:
    int mySqrt(int x) {
        if (x == 0) return 0;
        if (x == 1) return 1;
        int min = 0;
        int max = x;
        while (max - min > 1) {
            int m = (max + min)/2;
            if (x / m == m) return m;
            if (x / m < m) max = m;
            else min = m;
        }
        return min;
    }
};

[linyang4]

思路

二分法

代码

const mySqrt = x => {
    let left = 1, right = x
    let ans
    while(left <= right) {
        const mid = (left + right) >>> 1
        if (mid * mid === x) { return mid }
        if (mid * mid < x) { 
            ans = mid
            left++
        } else {
            right--
        }
    }
    return ans
}

复杂度

• 时间复杂度: O(logx)
• 空间复杂度: O(1)

[CodingProgrammer]

思路1

正常思路，i 从1累加，直到 i * i 等于 x ，返回 i; 大于 x ，返回 i - 1；小于 x ，i++

代码

class Solution {
    public int mySqrt(int x) {
        // 正常思路：从1开始一直累加计算乘积，如果等于 x，找到了；如果小于，继续累加1；如果大于，则超了，返回上一个值
        if (x == 0) return 0;
        int i = 1;
        while (i <= x) {
            int multi = i * i;
            if (multi > x || multi < 0) break;
            else if (multi < x) i++;
            else return i;
        }

        return i - 1;
    }
}

复杂度

• Time: O(N)
• Space: O(1)

[dahaiyidi]

Problem

69. Sqrt(x)

难度简单864

给你一个非负整数 x ，计算并返回 x 的 算术平方根 。

由于返回类型是整数，结果只保留 整数部分 ，小数部分将被 舍去 。

注意：不允许使用任何内置指数函数和算符，例如 pow(x, 0.5) 或者 x ** 0.5 。

---

Note

• 二分查找：时间logn, 空间1
• 牛顿迭代：时间logn，空间1

---

Complexity

• 时间O：
• 空间O：

---

Python

方法1：二分查找
class Solution:
    def mySqrt(self, x: int) -> int:
        l = 0
        r = x
        while l <= r:
            mid = l + ((r - l) >> 1)
            val = mid * mid
            if val < x:
                l = mid + 1
            elif val > x:
                r = mid - 1
            else:
                return mid
        # 最终退出时，l位于刚好大于目标值的位置上，r位于刚好小于目标值的位置上
        return r

方法2：牛顿迭代法
class Solution:
    def mySqrt(self, x: int) -> int:
        if x == 0:
            return 0
        # 牛顿迭代法
        x_old = x
        while 1:
            x_new = (x_old + x / x_old) / 2  # 根据梯度更新
            if abs(x_old - x_new) < 1e-7:
                break
            x_old = x_new  
        # 最终停下的点位于真实值的右侧
        return int(x_new)

C++

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

From : https://github.com/dahaiyidi/awsome-leetcode

[CodingProgrammer]

思路2

二分法，从1累加具有单调递增性质，考虑使用二分法

代码

class Solution {
    public int mySqrt(int x) {
        // 思路2: 二分法，从1累加具有单调递增性质，考虑使用二分法
        if (x < 2) return x;
        int left = 1, right = x;
        while (left + 1 < right) {
            int mid = left + ((right - left) >> 1);
            if (mid >= x / mid) {
                right = mid;
            } else {
                left = mid;
            }
        }

        return right > x / right ? left : right;
    }
}

复杂度 :

• Time: O(logN)
• Space: O(1)

[guangsizhongbin]

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

[L-SUI]

/**
 * @param {number} x
 * @return {number}
 */
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
}

[Alfie100]

LeetCode题目连接： 69. Sqrt(x)https://leetcode-cn.com/problems/sqrtx/

解题思路

二分法，得到二分结果后再判断其平方是否大于目标值 x。

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:
                return mid
            elif mid*mid < x:
                left = mid+1
            else:
                right = mid

        if left*left > x:
            return left-1
        else:
            return left

复杂度分析

• 时间复杂度：O(log x)。
• 空间复杂度：O(1)。

[LiFuQuan1208]

思路：

基本的二分法，和第27天的基本是一样的。 换个变量就好了

代码：

 public  static int mySqrt(int x) {
        if(x==1) return 1;
        int begin=0;
        int end=46340;
        while(begin<=end){
            int mid=begin+(end-begin)/2;
            if(mid*mid>x){
                end=mid-1;
            }else if(mid*mid<x){
                begin=mid+1;
            }else{
                return mid;
            }
        }
        return end;
    }

[machuangmr]

代码

class Solution {
    public int mySqrt(int x) {
        // 利用二分法
        int left = 0, right = x, ans = -1;
        while(left <= right) {
            int mid = left + (right - left) / 2;
            if((long)mid * mid <= x) {
                ans = mid;
                left = mid  + 1;
            } else {
                right = mid - 1;
            }
        }
        return ans;
    }
}

复杂度

• 时间复杂度 O(logx)
• 空间复杂度 O(1)

[Today-fine]

二分法查找

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

[Moin-Jer]

思路

---

二分

代码

---

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

复杂度分析

---

• 时间复杂度：O(logx)
• 空间复杂度：O(1)

[zzzpppy]

二分

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;
            if(mid > x / mid){

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

时间复杂度O(logN) 空间复杂度O(1)

[spacker-343]

思路

排除法

代码

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

复杂度

时间复杂度：O(log2n) 空间复杂度：O(n)

[1149004121]

69. Sqrt(x)

思路

二分法搜索，找到最大的res，使得res * res <= x。

代码

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

复杂度分析

• 时间复杂度：O(logN)。
• 空间复杂度：O(1)。

[tian-pengfei]

class Solution {
public:
    int mySqrt(int x) {
        if(x==1)return 1;
        long target =x;
        long low = 0;
        long mid = target/2;
        long high= target;
         while ((mid*mid!=target)&&!(mid*mid<x&&(mid+1)*(mid+1)>target)){

           if(mid*mid>x){
                high = mid-1;
            }else{
                low = mid+1;
            }
            mid = (low+high)/2;
        }
        return mid;
    }
};

[MissNanLan]

代码

• 语言支持：JavaScript

JavaScript Code:

var mySqrt = function (x) {
  let l = 0,
    r = x,
    ans = -1;
  while (l <= r) {
    const mid = Math.floor((l+r)/2)

    if (mid * mid <= x) {
      ans = mid;
      l = mid + 1;
    } else {
      r = mid - 1;
    }
  }
  return ans;
};

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)

[for123s]

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

[Jay214]

思路

二分法，无限向右逼近

代码

/**
 * @param {number} x
 * @return {number}
 */
var mySqrt = function(x) {
    if (x === 0 ) return 0;
    if (x < 4) return 1;
    let left = 0;
    let right = x;
    let middle;
    let ans
    while (left <= right) {
          middle = Math.floor(left + (right - left)/2);
        if (middle * middle <= x) {
            ans = middle
            left = middle + 1;
        } else {
            right = middle - 1;
        }
    }
    return ans;

};

复杂度分析

时间log(x)，空间O(1)

[codingPitaya]

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

• 时间复杂度 O(logx)
• 空间复杂度 O(1)

[BpointA]

思路

x的平方根

代码

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

        }
        return right;

    }
}

[chakochako]

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

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

[GaoMinghao]

思路

小于右边界的最大值

代码

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

复杂度

O(lgN)

[1916603886]

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 };