动态规划例子：Maximal Square

Given a 2D binary matrix filled with 0's and 1's, find the largest square containing all 1's and return its area.

For example, given the following matrix:

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Return 4.

 

 1 public class Solution {
 2     public int maximalSquare(char[][] matrix) {
 3         int rows = matrix.length;
 4         if(rows == 0) return 0;
 5         int cols = matrix[0].length;
 6         if(cols == 0) return 0;
 7         Node[][] sta = new Node[rows][cols];
 8         for(int i=0; i<rows; i++){
 9             for(int j=0; j<cols; j++){
10                 sta[i][j] = new Node();
11             }
12         }
13         
14         int res = 0;
15         
16         if(matrix[0][0] == '1'){
17             sta[0][0].left = 1;
18             sta[0][0].up = 1;
19             sta[0][0].maxSize = 1;
20             res = 1;
21         }
22         //求第一行
23         for(int i=1; i<cols; i++){
24             if(matrix[0][i] == '1'){
25                 sta[0][i].left = sta[0][i-1].left+1;
26                 sta[0][i].maxSize = 1;
27                 res = 1;
28             }
29         }
30         //求第一列
31         for(int j=1;j<rows;j++){
32             if(matrix[j][0] == '1'){
33                 sta[j][0].up = sta[j-1][0].up + 1;
34                 sta[j][0].maxSize = 1;
35                 res = 1;
36             }
37         }
38         //动态求其他
39         for(int i=1; i<rows; i++){
40             for(int j=1; j<cols; j++){
41                 if(matrix[i][j] == '1'){
42                     sta[i][j].left = sta[i-1][j].left + 1;
43                     sta[i][j].up = sta[i][j-1].up + 1;
44                     sta[i][j].maxSize = 1;
45                     if(matrix[i-1][j-1] == '1'){
46                         sta[i][j].maxSize = Math.min(sta[i][j].left, sta[i][j].up);
47                         sta[i][j].maxSize = Math.min(sta[i][j].maxSize, sta[i-1][j-1].maxSize+1);
48                     }
49                     
50                 }
51                 res = Math.max(sta[i][j].maxSize, res);
52             }
53         }
54         
55         return res*res;
56         
57     }
58     
59     class Node{
60         int left;
61         int up;
62         int maxSize;
63     }
64 
65 }