Set Matrix Zeroes

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.

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?

方法

空间复杂度要求为O(1)，肯定需要考虑如何使用数组本身，使用数组的第一行和第一列作为标记。

    public void setZeroes(int[][] matrix) {        if (!(matrix == null || matrix.length == 0 || matrix[0].length == 0)) {            int lenI = matrix.length;            int lenJ = matrix[0].length;            boolean rowFirst = false;            boolean colFirst = false;            for (int i = 0; i < lenI; i++) {                if (matrix[i][0] == 0) {                    colFirst = true;                    break;                }            }                        for (int j = 0; j < lenJ; j++) {                if (matrix[0][j] == 0) {                    rowFirst = true;                    break;                }            }                        for (int i = 0; i < lenI; i++) {                for (int j = 0; j < lenJ; j++) {                    if (matrix[i][j] == 0) {                        matrix[i][0] = 0;                        matrix[0][j] = 0;                    }                }            }            for (int i = 1; i < lenI; i++) {                if (matrix[i][0] == 0) {                    for (int j = 1; j < lenJ; j++) {                        matrix[i][j] = 0;                    }                }            }                        for (int j = 1; j < lenJ; j++) {                if (matrix[0][j] == 0) {                    for (int i = 1; i < lenI; i++) {                        matrix[i][j] = 0;                    }                }            }                        if (rowFirst) {                for (int j = 0; j < lenJ; j++) {                    matrix[0][j] = 0;                }            }                        if (colFirst) {                for (int i = 0; i < lenI; i++) {                    matrix[i][0] = 0;                }            }        }    }