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 }
