Leetcode--Maximal Square

Title Description:


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.




In a matrix, find the largest square area (1 is the point of the square, and 0 means empty).




The topic is a typical DP problem.


1. Assign a value of dp[0,i] to Matrix[0,i] and dp[i,0] to matrix[i,0]. I∈[0,n)


2. The two-layer loop is assigned a value of dp[i,j according to different conditions:
A. Matrix[i,j] = = 1 and 3 neighbors (dp[i-1,j],dp[i,j-1],dp[i-1,j-1]) are 1:dp[i,j] = 4
B. matrix[i,j] = = 1 and 3 neighbors >1 and equal: dp[i,j] = (square root of neighbor area +1) squared
C. matrix[i,j] = = 1 and 3 neighbors >=1 but not necessarily equal: dp[i,j]= (the square root of the smallest value in the neighbor +1) squared
D. Other circumstances: dp[i,j] = matrix[i,j]


3. Use the max variable to track the maximum value of the current DP[I,J]


Implementation code:

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) {m        AX = 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]);} Neighbours all no less than 1, and may not equals each otherelse if (matrix[i,j] = = ' 1 ' && dp[i-1,j] >= 1 &am p;& 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) + +);}                else{Dp[i,j] = matrix[i,j] = = ' 1 '? 1:0;            }if (Dp[i,j] > max) {max = dp[i,j];}            }} return max; }}

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Leetcode--Maximal Square