Set Matrix Zeroes-优快云博客

Set Matrix Zeroes

leetcode73

题目：

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

Example 1:

Input: 
[
  [1,1,1],
  [1,0,1],
  [1,1,1]
]
Output: 
[
  [1,0,1],
  [0,0,0],
  [1,0,1]
]

Example 2:

Input: 
[
  [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]
]

Follow up:

A straight forward solution using O(m**n) 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?

空间复杂度 O(m*n) 或 O(m+n) 比较容易实现，难度在于实现空间复杂度 O(1) 的算法。

思路：

空间复杂度 O(1) 代表我们不能创建行/列长度的状态数组，考虑借用输入数组存放状态变量。例如，选取第1列存放状态变量，在这之前先将第一列的状态保存下来。

class Solution {
    public void setZeroes(int[][] matrix) {
        // col0用来保存第一列状态
        int col0 = 1;
        int rows = matrix.length;
        int cols = matrix[0].length;
        for (int row = 0; row < rows; row++) {
            if(matrix[row][0] == 0)
                col0 = 0;
            for(int col = 1; col < cols; col++){
                // 将状态保存到第一列
                if(matrix[row][col] == 0){
                    matrix[row][0] = 0;
                    matrix[0][col] = 0;
                }
            }
        }

        for (int row = rows-1; row >= 0 ; row--) {
            for (int col = cols-1; col >= 1 ; col--) {
                if(matrix[0][col] == 0 || matrix[row][0] == 0)
                    matrix[row][col] = 0;
            }
            // 还原第一列状态
            if(col0 == 0)
                matrix[row][0] = 0;
        }
        return;
    }
}