问题描述
Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.
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?
code
/*
Contant Space，考虑定义几个变量或者加以题目本身的空间
本体利用矩阵的第一行和第一列来作为辅助空间使用，记录该行和该列有必要set to 0
同时注意第一行和第一列本身是否要set to 0
1.确定第一行和第一列是否需要清零
isFirstRowZero、isFirstColumnZero
2.扫描剩下的矩阵元素，如果遇到了0，就将对应的第一行和第一列上的元素赋值为0
matrix[i][0] = 0;
matrix[0][j] = 0;
3.根据第一行和第一列的信息，将该清0的清0
4.处理第一行和第一列
*/
public class Solution {
public void setZeroes(int[][] matrix) {
boolean isFirstRowZero = false;
boolean isFirstColumnZero = false;
for(int i = 0;i<matrix.length;i++) {
if(matrix[i][0] == 0) {
isFirstRowZero = true;
break;
}
}
for(int i = 0;i<matrix[0].length;i++) {
if(matrix[0][i] == 0) {
isFirstColumnZero = true;
break;
}
}
for(int i = 1;i<matrix.length;i++)
for(int j = 1;j<matrix[0].length;j++) {
if(matrix[i][j] == 0) {
matrix[i][0] = 0;
matrix[0][j] = 0;
}
}
for(int i = 1;i<matrix.length;i++) {
if(matrix[i][0] == 0) {
for(int j = 1;j<matrix[0].length;j++)
matrix[i][j] = 0;
}
}
for(int i = 1;i<matrix[0].length;i++) {
if(matrix[0][i] == 0) {
for(int j = 1;j<matrix.length;j++)
matrix[j][i] = 0;
}
}
if(isFirstRowZero)
for(int i = 0;i<matrix.length;i++)
matrix[i][0] = 0;
if(isFirstColumnZero)
for(int i = 0;i<matrix[0].length;i++)
matrix[0][i] = 0;
}
}