LeetCode | 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.

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?

分析
题目要求使用常数的额外空间，因此需要借助矩阵本身的空间来辅助存储，这里借用了矩阵的第一行和第一列来辅助纪录该行或该列是否为0.由于第一行第一列自身发生了改变，再用两个布尔变量纪录第一行和第一列是否为0即可。

代码

public class SetMatrixZeroes {
public void setZeroes(int[][] matrix) {
if (matrix == null || matrix.length == 0) {
return;
}
int M = matrix.length;
int N = matrix[0].length;
boolean isFirstRowZero = false;
boolean isFirstColumnZero = false;
for (int j = 0; j < N; ++j) {
if (matrix[0][j] == 0) {
isFirstRowZero = true;
break;
}
}
for (int i = 0; i < M; ++i) {
if (matrix[i][0] == 0) {
isFirstColumnZero = true;
break;
}
}
for (int i = 1; i < M; ++i) {
for (int j = 1; j < N; ++j) {
if (matrix[i][j] == 0) {
matrix[i][0] = 0;
matrix[0][j] = 0;
}
}
}

for (int i = 1; i < M; ++i) {
for (int j = 1; j < N; ++j) {
if (matrix[i][0] == 0 || matrix[0][j] == 0)
matrix[i][j] = 0;
}
}

if (isFirstRowZero) {
for (int j = 0; j < N; ++j) {
matrix[0][j] = 0;
}
}

if (isFirstColumnZero) {
for (int i = 0; i < M; ++i) {
matrix[i][0] = 0;
}
}
}
}