[LeetCode]Set Matrix Zeroes

Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.

in place，主要是感觉比较麻烦，其实就是利用第一行和第一列保存该行列是否需要变0，再用两个变量表示第一行和第一列是否需要变0

代码：

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

for(int i =1; i < m; ++ i){
if(matrix[i][0] == 0){
for (int j = 0; j < n; ++j){
matrix[i][j] = 0;
}
}
}
if(colZero){
for(int i = 0; i< m; ++i){
matrix[i][0] = 0;
}
}

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

}

}

此外我们可以记录相应的0的位置，我这其实是浪费空间了，其实可以优化到O（m+n）的空间

public void setZeroes1(int[][] matrix) {
int m = matrix.length, n = matrix[0].length;
class Node {
int x;
int y;

public Node(int x, int y) {
this.x = x;
this.y = y;
}
}
List<Node> nodes = new LinkedList<Node>();
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (matrix[i][j] == 0) {
nodes.add(new Node(i, j));
}
}
}
for (Node node : nodes) {
for (int i = 0; i < m; i++) {
matrix[i][node.y] = 0;
}
for (int j = 0; j < n; ++j) {
matrix[node.x][j] = 0;
}
}
}