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91 天学算法第五期打卡
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【Day 89 】2021-12-07 - 378. 有序矩阵中第 K 小的元素 #108

Open azl397985856 opened 2 years ago

azl397985856 commented 2 years ago

378. 有序矩阵中第 K 小的元素

入选理由

暂无

题目地址

https://leetcode-cn.com/problems/kth-smallest-element-in-a-sorted-matrix/

前置知识

 

示例：

matrix = [ [ 1, 5, 9], [10, 11, 13], [12, 13, 15] ], k = 8,

返回 13。  

提示： 你可以假设 k 的值永远是有效的，1 ≤ k ≤ n2 。

chaggle commented 2 years ago

title: "Day 89 378. 有序矩阵中第 K 小的元素" date: 2021-12-07T08:38:44+08:00 tags: ["Leetcode", "c++", "unordered_map"] categories: ["91-day-algorithm"] draft: true

378. 有序矩阵中第 K 小的元素

题目

给你一个 n x n 矩阵 matrix ，其中每行和每列元素均按升序排序，找到矩阵中第 k 小的元素。
请注意，它是 排序后 的第 k 小元素，而不是第 k 个 不同 的元素。

 

示例 1：

输入：matrix = [[1,5,9],[10,11,13],[12,13,15]], k = 8
输出：13
解释：矩阵中的元素为 [1,5,9,10,11,12,13,13,15]，第 8 小元素是 13
示例 2：

输入：matrix = [[-5]], k = 1
输出：-5
 

提示：

n == matrix.length
n == matrix[i].length
1 <= n <= 300
-109 <= matrix[i][j] <= 109
题目数据 保证 matrix 中的所有行和列都按 非递减顺序 排列
1 <= k <= n^2

题目思路

  • 1、题目中最合适的办法理应是二分的思想。
class Solution {
public:
    bool check(vector<vector<int>>& matrix, int mid, int k, int n) {
        int i = n - 1, j = 0, num = 0;
        while (i >= 0 && j < n) 
        {
            if (matrix[i][j] <= mid) 
            {
                num += i + 1;
                j++;
            } else i--;
        }
        return num >= k;
    }

    int kthSmallest(vector<vector<int>>& matrix, int k) {
        int n = matrix.size();
        int l = matrix[0][0], r = matrix[n - 1][n - 1];
        while(l < r) 
        {
            int mid = l + ((r - l) >> 1);
            if (check(matrix, mid, k, n)) r = mid;
            else l = mid + 1;
        }
        return l;
    }
};

复杂度

machuangmr commented 2 years ago

代码

class Solution {
    public int kthSmallest(int[][] matrix, int k) {
        int n = matrix.length;
        int left = matrix[0][0];
        int right = matrix[n - 1][n - 1];
        while (left < right) {
            int mid = left + ((right - left) >> 1);
            if (check(matrix, mid, k, n)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }

    public boolean check(int[][] matrix, int mid, int k, int n) {
        int i = n - 1;
        int j = 0;
        int num = 0;
        while (i >= 0 && j < n) {
            if (matrix[i][j] <= mid) {
                num += i + 1;
                j++;
            } else {
                i--;
            }
        }
        return num >= k;
    }
}

复杂度

ymkymk commented 2 years ago

优先队列

class Solution { public int kthSmallest(int[][] matrix, int k) { PriorityQueue<int[]> pq = new PriorityQueue<int[]>(new Comparator<int[]>() { public int compare(int[] a, int[] b) { return a[0] - b[0]; } }); int n = matrix.length; for (int i = 0; i < n; i++) { pq.offer(new int[]{matrix[i][0], i, 0}); } for (int i = 0; i < k - 1; i++) { int[] now = pq.poll(); if (now[2] != n - 1) { pq.offer(new int[]{matrix[now[1]][now[2] + 1], now[1], now[2] + 1}); } } return pq.poll()[0]; } }

时间复杂度：O(klogn)，归并 k 次，每次堆中插入和弹出的操作时间复杂度均为logn。

空间复杂度：O(n)，堆的大小始终为 n。

Zhang6260 commented 2 years ago

JAVA版本

思路：

主要使用二分的思想

class Solution {
    public int kthSmallest(int[][] matrix, int k) {
        int n = matrix.length;
        int left = matrix[0][0];
        int right = matrix[n - 1][n - 1];
        while (left < right) {
            int mid = left + ((right - left) >> 1);
            if (check(matrix, mid, k, n)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }

    public boolean check(int[][] matrix, int mid, int k, int n) {
        int i = n - 1;
        int j = 0;
        int num = 0;
        while (i >= 0 && j < n) {
            if (matrix[i][j] <= mid) {
                num += i + 1;
                j++;
            } else {
                i--;
            }
        }
        return num >= k;
    }
}

时间复杂度：O(logn)

空间复杂度：O(1)

mokrs commented 2 years ago 
int kthSmallest(vector<vector<int>>& matrix, int k) {
    priority_queue<int, vector<int>, greater<int>> q;

    for (int i = 0; i < matrix.size(); ++i)
    {
        for (int j = 0; j < matrix.front().size(); ++j)
        {
            q.emplace(matrix[i][j]);
        }
    }

    for (; k > 1; --k){
        q.pop();
    }

    return q.top();
}
yibenxiao commented 2 years ago

【Day 89】378. 有序矩阵中第 K 小的元素

代码

class Solution {
public:
    bool check(vector<vector<int>>& matrix, int mid, int k, int n) {
        int i = n - 1;
        int j = 0;
        int num = 0;
        while (i >= 0 && j < n) {
            if (matrix[i][j] <= mid) {
                num += i + 1;
                j++;
            } else {
                i--;
            }
        }
        return num >= k;
    }

    int kthSmallest(vector<vector<int>>& matrix, int k) {
        int n = matrix.size();
        int left = matrix[0][0];
        int right = matrix[n - 1][n - 1];
        while (left < right) {
            int mid = left + ((right - left) >> 1);
            if (check(matrix, mid, k, n)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
};

复杂度

时间复杂度：O(NLogN)

空间复杂度：O(1)

wangyifan2018 commented 2 years ago 
class Solution:
    def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
        n = len(matrix)
        def check(mid):
            i, j = n - 1, 0
            num = 0
            while i >= 0 and j < n:
                if matrix[i][j] <= mid:
                    num += i + 1
                    j += 1
                else:
                    i -= 1
            return num >= k
        left, right = matrix[0][0], matrix[-1][-1]
        while left < right:
            mid = (left + right) // 2
            if check(mid):
                right = mid
            else:
                left = mid + 1
        return left
BreezePython commented 2 years ago

思路

二分查找

代码

class Solution:
    def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
        n = len(matrix)

        def check(mid):
            i, j = n - 1, 0
            num = 0
            while i >= 0 and j < n:
                if matrix[i][j] <= mid:
                    num += i + 1
                    j += 1
                else:
                    i -= 1
            return num >= k

        left, right = matrix[0][0], matrix[-1][-1]
        while left < right:
            mid = (left + right) // 2
            if check(mid):
                right = mid
            else:
                left = mid + 1

        return left

复杂度

kidexp commented 2 years ago

thoughts

相当于多列list 求第k小个数 就是用heap 存每一行的head

code

import heapq

class Solution:
    def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
        """
        heap
        """
        heap = [(array[0], index, 0) for index, array in enumerate(matrix)]
        heapq.heapify(heap)
        for _ in range(k):
            num, row, col = heapq.heappop(heap)
            if col < len(matrix[row]) - 1:
                heapq.heappush(heap, (matrix[row][col + 1], row, col + 1))
        return num

Complexity

时间复杂度O(n +klgn)

空间复杂度O(n)

Richard-LYF commented 2 years ago

class Solution: def kthSmallest(self, matrix: List[List[int]], k: int) -> int: n = len(matrix)

    def check(mid):
        i, j = n - 1, 0
        num = 0
        while i >= 0 and j < n:
            if matrix[i][j] <= mid:
                num += i + 1
                j += 1
            else:
                i -= 1
        return num >= k

    left, right = matrix[0][0], matrix[-1][-1]
    while left < right:
        mid = (left + right) // 2
        if check(mid):
            right = mid
        else:
            left = mid + 1

    return left
ysy0707 commented 2 years ago

思路：二分

class Solution {
    //2.二分查找
    public int kthSmallest(int[][] matrix, int k) {
         int n = matrix.length;
         int left = matrix[0][0];
         int right = matrix[n - 1][n - 1];

         //循环条件是 left < right 因为第k小元素一定存在，
         //所以当left == right时就是找到了想要的值了；如果是left <= right的话循环是无法终止的
         while(left < right){
            int mid = left + (right - left) / 2;
            //若不大于中间值mid的数大于等于k，说明mid可能取大了，因此令right = mid
            if(getCount(mid, n, matrix) >= k){
                right = mid;
            }
            //若不大于中间值mid的数比k少，说明mid取小了，令left = mid + 1
            else{
                left = mid + 1;
            }
        }
        return left;
    }
    //有点像剑指offer.04和240 从左下角开始找
    public int getCount(int mid, int n, int[][] matrix){
        int i = n - 1, j = 0;
        int count = 0;
        while(i >= 0 && j < n){
            //当这个点小于mid 说明这个点同列以上的所有点都小于mid，可以计数,同时列向右加一
            if(matrix[i][j] <= mid){
                count += i + 1;
                j++;
            }
            //如果大于了mid,这一行右侧的都大于mid,上移一行
            else{
                i--;
            }
        }
        return count;
    }
}

时间复杂度：O(NLogN) 空间复杂度：O(1)

winterdogdog commented 2 years ago 
bool check(int **matrix, int mid, int k, int n) {
    int i = n - 1;
    int j = 0;
    int num = 0;
    while (i >= 0 && j < n) {
        if (matrix[i][j] <= mid) {
            num += i + 1;
            j++;
        } else {
            i--;
        }
    }
    return num >= k;
}

int kthSmallest(int **matrix, int matrixSize, int *matrixColSize, int k) {
    int left = matrix[0][0];
    int right = matrix[matrixSize - 1][matrixSize - 1];
    while (left < right) {
        int mid = left + ((right - left) >> 1);
        if (check(matrix, mid, k, matrixSize)) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
}
joriscai commented 2 years ago

思路

代码

javascript

/*
 * @lc app=leetcode.cn id=378 lang=javascript
 *
 * [378] 有序矩阵中第 K 小的元素
 */

// @lc code=start
/**
 * @param {number[][]} matrix
 * @param {number} k
 * @return {number}
 */
var kthSmallest = function(matrix, k) {
  const n = matrix.length
  let low = matrix[0][0]
  let high = matrix[n - 1][n - 1]
  while (low <= high) {
    let midVal = low + ((high - low) >>> 1)   // 获取中间值
    let count = countInMatrix(matrix, midVal) // 找出矩阵中小于等于它的个数
    if (count < k) {
      low = midVal + 1
    } else {
      high = midVal - 1
    }
  }
  return low
};

const countInMatrix = (matrix, midVal) => {
  const n = matrix.length             // 矩阵是 n行n列
  let count = 0
  let row = 0                         // 第一行
  let col = n - 1                     // 最后一列
  while (row < n && col >= 0) {
    if (midVal >= matrix[row][col]) { // 大于等于当前行的最右
      count += col + 1                // 不大于它的数增加col + 1个
      row++                           // 比较下一行
    } else {                          // 干不过当前行的最右元素
      col--                           // 留在当前行，比较左边一个
    }
  }
  return count
};
// @lc code=end

复杂度分析

vincentLW commented 2 years ago 
import heapq
class Solution:
    def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
        l = []
        element = -1
        for i in range(len(matrix)):
            heapq.heappush(l, (matrix[i][0], i, 0))
        for i in range(k):
            element, row, column = heapq.heappop(l)
            if column + 1 < len(matrix[row]): heapq.heappush(l, (matrix[row][column + 1], row, column + 1))
        return element

时间复杂度：O(n * log(r - l))，二分查找进行次数为O(log(r−l))，每次操作时间复杂度为O(n)。 空间复杂度：O(1)，常数空间

ZETAVI commented 2 years ago 
class Solution {
public:
    bool check(vector<vector<int>>& m, int k, int n, int mid) {
        int i = n - 1;
        int j = 0;
        int cnt = 0;
        while (i >= 0 && j < n) {
            if (m[i][j] <= mid) {
                cnt += i + 1;
                j++;
            } else {
                i--;
            }
        }
        return cnt >= k;
    }
    int kthSmallest(vector<vector<int>>& matrix, int k) {
        int n = matrix.size();
        int l = matrix[0][0];
        int r = matrix[n - 1][n - 1];
        while (l < r) {
            int mid = l + (r - l) / 2;
            if (check(matrix, k, n, mid)) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
};
yj9676 commented 2 years ago 
class Solution {
    public int kthSmallest(int[][] matrix, int k) {
        int rows = matrix.length, columns = matrix[0].length;
        int[] sorted = new int[rows * columns];
        int index = 0;
        for (int[] row : matrix) {
            for (int num : row) {
                sorted[index++] = num;
            }
        }
        Arrays.sort(sorted);
        return sorted[k - 1];
    }
}
taojin1992 commented 2 years ago 
# Understand:
Given an n x n matrix where each of the rows and columns are sorted in ascending order

it is the kth smallest element in the sorted order, not the kth distinct element

n == matrix.length
n == matrix[i].length
1 <= n <= 300
-10^9 <= matrix[i][j] <= 10^9
All the rows and columns of matrix are guaranteed to be sorted in non-decreasing order.
1 <= k <= n^2, k is always valid

k is 1-indexed

matrix = 
[[1,5,9],
[10,11,13],
[12,13,15]]
-> 13

matrix = [[-5]], k = 1
-> -5

# Plan & Match:

# Bruteforce:
add all to minheap, poll k times
O(n^2 * log(n ^ 2)) + O(klog(n^2))
-> O(n^2 * log(n ^ 2))

# Optimization:
add head to each row into min heap, poll() once (in total we poll k times), add 1 element from the same row
corner case: a row can be used up earlier 

How to have n pointers to each row? 
when insert into the heap, insert an object, [value, rowIndex, colIndex]
rewrite the comparator

Code:


class Solution {
    class Wrapper {
        int val;
        int rowIndex;
        int colIndex;

        Wrapper(int val, int rowIndex, int colIndex) {
            this.val = val;
            this.rowIndex = rowIndex;
            this.colIndex = colIndex;
        }
    }

    public int kthSmallest(int[][] matrix, int k) {
        // edge case
        if (matrix.length == 1) return matrix[0][0];
        int n = matrix.length;
        PriorityQueue<Wrapper> minHeap = new PriorityQueue<>((a, b) -> (a.val - b.val));
        // pre-processing
        for (int row = 0; row < n; row++) {
            minHeap.offer(new Wrapper(matrix[row][0], row, 0));
        }
        while (k > 1) {
            k -= 1;
            Wrapper cur = minHeap.poll();
            int curRow = cur.rowIndex;
            int curCol = cur.colIndex;
            // edge case, nothing to use on the same row
            if (curCol == n - 1) {
                continue;
            } else {
                minHeap.offer(new Wrapper(matrix[curRow][curCol + 1], curRow, curCol + 1));
            }
        }
        return minHeap.poll().val;                                                   
    }
}
ChenJingjing85 commented 2 years ago 
class Solution:
    def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
        l = []
        element = -1
        for i in range(len(matrix)):
            heapq.heappush(l, (matrix[i][0], i, 0))
        for i in range(k):
            element, row, column = heapq.heappop(l)
            if column + 1 < len(matrix[row]): heapq.heappush(l, (matrix[row][column + 1], row, column + 1))
        return element
for123s commented 2 years ago

代码

C++ Code:


class Solution {
public:
    bool findnum(vector<vector<int>> matrix,int n,int mid,int k)
    {
        int i = 0, j = n - 1;
        int ans = 0;
        while(i<n&&j>=0)
        {
            if(matrix[j][i]>mid)
            {    
                --j;
            }
            else
            {    
                ++i;
                ans += j+1;
            }
        }
        return ans >= k;
    }

    int kthSmallest(vector<vector<int>>& matrix, int k) {
        int n = matrix.size();
        int left = matrix[0][0], right = matrix[n-1][n-1];
        while(left<right)
        {
            int mid = left+((right-left)>>1);
            if(findnum(matrix, n, mid, k))
                right = mid;
            else
                left = mid+1;
        }
        return left;
    }
};
ForLittleBeauty commented 2 years ago

思路

Basic idea is binary search in 2D matrix.

  1. find mid value
  2. count number of element that is less tha mid value
  3. if the number < k : left = mid+1. if the number>=k: right = mid-1
  4. when left == right, it is on the right number

代码

class Solution:
    def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
        N = len(matrix)
        left, right = matrix[0][0], matrix[N - 1][N -1]

        def less_than(mid):
            count = 0
            row, col = N - 1, 0
            while row >= 0 and col < N:
                if matrix[row][col] <= mid:
                    count += row + 1
                    col += 1                    
                else:
                    row -= 1
            return count

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

时间复杂度: O(nlog(r-l))

空间复杂度: O(1)

Bingbinxu commented 2 years ago

思路 先构建一大顶堆，然后取出k小元素 代码(C++)

实现语言: C++
class Solution {
public:
    int kthSmallest(vector<vector<int>>& M, int k) {
        priority_queue<int, vector<int>, greater<int>> q;
        int res;        
        const int rows = M.size();
        const int cols = M.front().size(); 
        for (int i = 0; i < rows; i++)
        {
            for (int j = 0; j < cols; j++)
            {
                q.push(M[i][j]);
            }
        }
        while (!q.empty() && k - 1 > 0)  
        {
            q.pop();
            k--;
        }
        res = q.empty() ? -1 : q.top();
        return res;
    }
};

复杂度分析 时间复杂度: O(NlogN) 空间复杂度: O(N）