leetcode73 - Set Matrix Zeroes - medium

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

Follow up:

• 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?

 

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

 

实时改动的话，比如例1里把第一行第一列直接set成0，那接下来我们会遍历到一个被改成0的位置，再实时改一下，全盘都要变0.

所以我们用第0行和第0列来做标记，第一次遍历，碰到m[i][j]是0的话，把m[i][0]和m[0][j]标成0，表示等会儿第i行和第j列是要动的。

corner case：如果在第0行第5列有个0，用上面的办法m[0][0]会被标记，那第0列也受影响了。因此第0行和第0列单独check，并且用两个var来标记。之后第0行第0列也在最后再改动成0。

 

实现：Time O(mn) Space O(1)

class Solution {
public:
    void setZeroes(vector<vector<int>>& matrix) {
        
        bool firstRowZero = false, firstColZero = false;
        int m = matrix.size(), n = matrix[0].size();
        for (int i=0; i<m; i++){
            if (matrix[i][0] == 0){
                firstColZero = true;
                break;
            }
        }
        for (int i=0; i<n; i++){
            if (matrix[0][i] == 0){
                firstRowZero = true;
                break;
            }
        }
        
        for (int i=1; i<m; i++){
            for (int j=1; j<n; j++){
                if (matrix[i][j] == 0){
                    matrix[i][0] = 0;
                    matrix[0][j] = 0;
                }
            }
        }
        
        for (int i=1; i<m; i++){
            if (matrix[i][0] == 0){
                for (int j=0; j<n; j++){
                    matrix[i][j] = 0;
                }
            }
        }
        
        for (int j=1; j<n; j++){
            if (matrix[0][j] == 0){
                for (int i=0; i<m; i++){
                    matrix[i][j] = 0;
                }
            }
        }
        
        if (firstRowZero){
            for (int i=0; i<n; i++){
                matrix[0][i] = 0;
            }
        }
        if (firstColZero){
            for (int i=0; i<m; i++){
                matrix[i][0] = 0;
            }
        }
        
    }
};