题目描述：
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.
在一个矩阵中，找到最大的正方形面积（1表示正方形的点，0表示空），
LeetCode Maximal Square
。本题是很典型的DP问题。
1. 把dp[0,i]赋值为matrix[0,i] , dp[i,0]赋值为matrix[i,0]。i∈[0,n)
2. 两层循环根据不同情况为dp[i,j]赋值：
a. matrix[i,j] == 1 且3个邻居(dp[i-1,j],dp[i,j-1],dp[i-1,j-1])均为1 : dp[i,j] = 4
b. matrix[i,j] == 1 且3个邻居>1且相等: dp[i,j] = （邻居面积的平方根+1）的平方
c. matrix[i,j] == 1 且3个邻居>=1但不一定相等: dp[i,j]=（邻居中最小值的平方根+1）的平方
d. 其他情况： dp[i,j] = matrix[i,j]
3.使用max变量来track当前dp[i,j]的最值
实现代码：
public class Solution { public int MaximalSquare(char[,] matrix) { var row = matrix.GetLength(0); var col = matrix.GetLength(1); if(row < 2){ if(row == 0){ return 0; } else if(col == 1){ return matrix[0,0] == '1' ? 1 : 0; } } var max = 0; var dp = new int[row, col]; for(var i = 0;i < row; i++){ var x = matrix[i,0] == '1' ? 1 : 0; dp[i, 0] = x; if(dp[i,0] > max){max = dp[i,0];} } for(var i = 0;i < col; i++){ var x = matrix[0,i] == '1' ? 1 : 0; dp[0, i] = x; if(dp[0,i] > max){max = dp[0,i];} } for(var i = 1;i < row; i++){ for(var j = 1;j < col; j++){ // neighbours all equals 1 if(matrix[i,j] == '1' && dp[i-1,j] == 1 && dp[i, j-1] == 1 && dp[i-1,j-1] == 1){ if(dp[i-1, j] == 1){ dp[i,j] = 4; } } // neighbours all bigger than 1 and equals each other else if(matrix[i,j] == '1' && dp[i-1,j] == dp[i,j-1] && dp[i-1,j-1] == dp[i-1,j] && dp[i-1,j] > 1){ dp[i,j] = (int)Math.Pow(Math.Sqrt(dp[i,j-1]) + 1,2); } // neighbours all no less than 1, but may not equals each other else if(matrix[i,j] == '1' && dp[i-1,j] >= 1 && dp[i,j-1] >= 1 && dp[i-1,j-1] >= 1){ var min = Math.Min(Math.Min(dp[i-1,j-1], dp[i-1,j]), dp[i,j-1]); dp[i,j] = (int)Math.Pow(Math.Sqrt(min) + 1,2); } else{ dp[i,j] = matrix[i,j] == '1' ? 1 : 0; } if(dp[i,j] > max){max = dp[i,j];} } } return max; }}