Leetcode: Set Matrix Zeroes

本文介绍了一种高效的算法,用于在原地将矩阵中所有值为零的元素对应的行和列置零。通过使用矩阵的第一行和第一列来存储状态信息,实现了常数空间复杂度的解决方案。

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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.

click to show follow up.

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?

Use the first row and the first column to store the status of that column or row. And use two flags to check whether the first row and column should be all set to zero. O(1) space and O(m * n) time.

public class Solution {
    public void setZeroes(int[][] matrix) {
        if(matrix==null || matrix.length==0 || matrix[0].length==0)  
            return;  
        
        boolean rowZero = false;
        boolean colZero = false;
        for (int i = 0; i < matrix[0].length; i++) {
            if (matrix[0][i] == 0) {
                rowZero = true;
            }
        }
        
        for (int i = 0; i < matrix.length; i++) {
            if (matrix[i][0] == 0) {
                colZero = true;
            }
        }
        
        for (int i = 1; i < matrix.length; i++) {
            for (int j = 1; j < matrix[0].length; j++) {
                if (matrix[i][j] == 0) {
                    matrix[0][j] = 0;
                    matrix[i][0] = 0;
                }
            }
        }
        
        for (int i = 1; i < matrix.length; i++) {
            for (int j = 1; j < matrix[0].length; j++) {
                if (matrix[i][0] == 0 || matrix[0][j] == 0) {
                    matrix[i][j] = 0;
                }
            }
        }
        
        if (rowZero) {
            for (int i = 0; i < matrix[0].length; i++) {
                matrix[0][i] = 0;
            }
        }
        
        if (colZero) {
            for (int i = 0; i < matrix.length; i++) {
                matrix[i][0] = 0;
            }
        }
    }
}


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