LEETCODE: Set Matrix Zeroes-优快云博客

本文提供了一种高效的算法，在不使用额外空间的情况下，将矩阵中某个元素为零的整行和整列置零。通过利用矩阵本身来记录需要置零的行和列，该算法实现了常数空间复杂度。

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Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.

Follow up:

Did you use extra space?
A straight forward 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?

思路，找到一个0，用它对应的行和列作为记录该行或列是否要为0的空间。

class Solution {
public:
    void setZeroes(vector<vector<int> > &matrix) {
        int row = -1;
        int col = -1;
        // Find the one element which is 0, and use its row and column for recording.
        for(int ii = 0; ii < matrix.size(); ii ++) {
            for(int jj = 0; jj < matrix[0].size();jj ++) {
                if(matrix[ii][jj] == 0) {
                    row = ii; 
                    col = jj;
                    break;
                }
            }
        }
        
        if(row == -1/*&& col == -1*/) return;
        
        //Do recording
        for(int ii = 0; ii < matrix.size(); ii ++) {
            for(int jj = 0; jj < matrix[0].size(); jj ++) {
                if(matrix[ii][jj] == 0) {
                    matrix[row][jj] = 0;
                    matrix[ii][col] = 0;
                }
            }
        }
        
        for(int ii = 0; ii < matrix.size(); ii ++) {
            if(ii != row && matrix[ii][col] == 0) {
                for(int jj = 0; jj < matrix[0].size(); jj ++) {
                    matrix[ii][jj] = 0;
                }
            }
        }
        
        for(int ii = 0; ii < matrix[0].size(); ii ++) {
            if(ii != col && matrix[row][ii] == 0) {
                for(int jj = 0; jj < matrix.size(); jj ++) {
                    matrix[jj][ii] = 0;
                }
            }
        }
        
        for(int ii = 0; ii < matrix.size(); ii ++) matrix[ii][col] = 0;
        for(int ii = 0; ii < matrix[0].size(); ii ++) matrix[row][ii] = 0;
    }
};