LeetCode -- Maximal Square

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）。




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 otherelse 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 otherelse 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;            }}