73. 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.

链接： http://leetcode.com/problems/set-matrix-zeroes/

5/28/2017

3ms, 17%

先确定第一行和第一列是否应该全部置为0，用firstRow, firstColumn表示。这里不改数组的原因是这一行一列会被用作其他行列的标志位。

遍历其他行列，并将第一行第一列相应元素设为0

遍历第一行第一列将相应元素设为0

 1 public class Solution {
 2     public void setZeroes(int[][] matrix) {
 3         if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
 4             return;
 5         }
 6         int row = matrix.length;
 7         int column = matrix[0].length;
 8         boolean firstRow = false, firstColumn = false;
 9         
10         if (!firstColumn) {
11             for (int i = 0; i < row; i++) {
12                 if (matrix[i][0] == 0) {
13                     firstColumn = true;
14                     break;
15                 }
16             }
17         }
18         if (!firstRow) {
19             for (int i = 0; i < column; i++) {
20                 if (matrix[0][i] == 0) {
21                     firstRow = true;
22                     break;
23                 }
24             }
25         }        
26         for (int i = 1; i < row; i++) {
27             for (int j = 1; j < column; j++) {
28                 if (matrix[i][j] == 0) {
29                     matrix[0][j] = 0;
30                     matrix[i][0] = 0;
31                 }
32             }
33         }
34         for (int i = 1; i < row; i++) {
35             if (matrix[i][0] == 0) {
36                 for (int j = 1; j < column; j++) {
37                     matrix[i][j] = 0;
38                 }
39             }
40         }
41         for (int i = 1; i < column; i++) {
42             if (matrix[0][i] == 0) {
43                 for (int j = 1; j < row; j++) {
44                     matrix[j][i] = 0;
45                 }
46             }
47         }
48         if (firstColumn) {
49             for (int i = 0; i < row; i++) {
50                 matrix[i][0] = 0;
51             }
52         }
53         if (firstRow) {
54             for (int i = 0; i < column; i++) {
55                 matrix[0][i] = 0;
56             }
57         }
58         return;
59     }
60 }

别人的算法，大部分跟我写的一样。有一个特别精简的：设第一行第一列用top-down，设其他元素用bottom-up

https://discuss.leetcode.com/topic/5056/any-shorter-o-1-space-solution

 1 void setZeroes(vector<vector<int> > &matrix) {
 2     int col0 = 1, rows = matrix.size(), cols = matrix[0].size();
 3 
 4     for (int i = 0; i < rows; i++) {
 5         if (matrix[i][0] == 0) col0 = 0;
 6         for (int j = 1; j < cols; j++)
 7             if (matrix[i][j] == 0)
 8                 matrix[i][0] = matrix[0][j] = 0;
 9     }
10 
11     for (int i = rows - 1; i >= 0; i--) {
12         for (int j = cols - 1; j >= 1; j--)
13             if (matrix[i][0] == 0 || matrix[0][j] == 0)
14                 matrix[i][j] = 0;
15         if (col0 == 0) matrix[i][0] = 0;
16     }
17 }