【Leetcode】Set Matrix Zeroes

题目链接：https://leetcode.com/problems/set-matrix-zeroes/

题目：

m x n

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?

思路：

用上、左边界标记除了两边界外数组中哪一行哪一列是否有零，然后按标记结果处理除两边界外数组，最后处理两边界。空间复杂度O(1)，时间复杂度为O(m*n)。

1、判断数组左上边界是否有零
2、扫描除边界外数组元素，若有零在边界处标记
3、根据标记按行列处理边界外元素为0
4、若边界有零，则边界归0

算法：

public void setZeroes(int[][] matrix) {
		boolean left = false;
		boolean up = false;
		for (int i = 0; i < matrix.length; i++) { // 左边界是否有零
			if (matrix[i][0] == 0) {
				left = true;
			}
		}
		for (int i = 0; i < matrix[0].length; i++) {// 上边界是否有零
			if (matrix[0][i] == 0) {
				up = true;
			}
		}
		for (int i = 1; i < matrix.length; i++) {
			for (int j = 1; j < matrix[0].length; j++) {
				if (matrix[i][j] == 0) {// 在边界处标记
					matrix[i][0] = 0;
					matrix[0][j] = 0;
				}
			}
		}
		for (int i = 1; i < matrix.length; i++) {// 根据左边界标记按行处理标记外数组
			if (matrix[i][0] == 0) {
				for (int j = 0; j < matrix[0].length; j++) {
					matrix[i][j] = 0;
				}
			}
		}
		for (int i = 1; i < matrix[0].length; i++) {// 根据上边界标记按列处理标记外数组
			if (matrix[0][i] == 0) {
				for (int j = 0; j < matrix.length; j++) {
					matrix[j][i] = 0;
				}
			}
		}
		if (left) {// 处理边界
			for (int i = 0; i < matrix.length; i++) {
				matrix[i][0] = 0;
			}
		}
		if (up) {
			for (int i = 0; i < matrix[0].length; i++) {
				matrix[0][i] = 0;
			}
		}
	}