一、题目

73. Set Matrix Zeroes

Given an m x n integer matrix matrix, if an element is 0, set its entire row and column to 0’s.

You must do it in place.

Example 1:

Input: matrix = [[1,1,1],[1,0,1],[1,1,1]]
Output: [[1,0,1],[0,0,0],[1,0,1]]
Example 2:

Input: matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]
Output: [[0,0,0,0],[0,4,5,0],[0,3,1,0]]

Constraints:

m == matrix.length
n == matrix[0].length
1 <= m, n <= 200
-231 <= matrix[i][j] <= 231 - 1

Follow up:

A straightforward solution using O(mn) space is probably a bad idea.
A simple improvement uses O(m + n) space, but still not the best solution.
Could you devise a constant space solution?

题目来源： leetcode

二、题解

class Solution {
public:void fillZero(vector<vector<int>>& matrix,int row,int col){int m = matrix.size();int n = matrix[0].size();//将行置为0for(int j = 0;j < n;j++){matrix[row][j] = 0;}//将列置为0for(int i = 0;i < m;i++){matrix[i][col] = 0;}}void setZeroes(vector<vector<int>>& matrix) {int m = matrix.size();int n = matrix[0].size();vector<pair<int,int>> rowCols;//存储值为0的元素对应的行和列for(int i = 0;i < m;i++){for(int j = 0;j < n;j++){if(matrix[i][j] == 0) rowCols.push_back({i,j});}}//将对应的行和列填充为0for(int i = 0;i < rowCols.size();i++){fillZero(matrix,rowCols[i].first,rowCols[i].second);}}
};