题目描述

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?

代码

    public static void setZeroes(int[][] matrix) {

        if (matrix == null || matrix.length == 0) {
            return;
        }

        int mx = matrix.length;
        int my = matrix[0].length;

        // 两个变量，判断第一行和第一列是否有0
        boolean xflag = false, yflag = false;

        // 判断第一行是否有0
        for (int i = 0; i < mx; i++) {
            if (matrix[i][0] == 0) {
                xflag = true;
                break;
            }
        }

        // 判断第一列是否有0
        for (int i = 0; i < my; i++) {
            if (matrix[0][i] == 0) {
                yflag = true;
                break;
            }
        }

        // 其它行、列是否有0
        for (int i = 1; i < mx; i++) {
            for (int j = 1; j < my; j++) {
                if (matrix[i][j] == 0) {
                    matrix[i][0] = 0;
                    matrix[0][j] = 0;
                }
            }
        }

        // 对于第一列，为0，则将所在行变成0
        for (int i = 1; i < mx; i++) {
            if (matrix[i][0] == 0) {
                for (int j = 0; j < my; j++) {
                    matrix[i][j] = 0;
                }
            }
        }

        // 对于第一行，为0，则将所在列变成0
        for (int i = 0; i < my; i++) {
            if (matrix[0][i] == 0) {
                for (int j = 0; j < mx; j++) {
                    matrix[j][i] = 0;
                }
            }
        }

        // 若原来第一行中有0，则将整行置0
        if (xflag) {
            for (int i = 0; i < mx; i++) {
                matrix[i][0] = 0;
            }
        }

        // 若原来第一列中有0，则将整列置0
        if (yflag) {
            for (int i = 0; i < my; i++) {
                matrix[0][i] = 0;
            }
        }

    }