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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 0Return 4.
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Special thanks to@Freezenfor adding this problem and creating all test cases.
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Dynamic Programming
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(H) Maximal Rectangle
思路:
1.动态规划的题,可以维护一个ret[m*n]的整数数组去表示matrix各元素的状态;
2.首先边界i=0,j=0:第一行和第一列与matrix相同(即ret[i][0] = matrix[i][0],ret[0][i] = matrix[0][i]);
3.当i>0,j>0时,
如果matrix[i][j]=='0',ret[i][j]=0;
如果matrix[i][j]=='1',ret[i][j]由左上角的3个值推出;
推导公式,即状态转移方程为:ret[i][j]=min(ret[i-1][j],ret[i][j-1],ret[i-1][j-1]) + 1;
code:
class Solution { public: int maximalSquare(vector>& matrix) { int m = matrix.size(); if(m==0) return 0; int n = matrix[0].size(); vector> ret(m, vector(n,0)); int maxSize = 0; for(int i=0;i
当然还可以有一些空间上的优化:Easy DP solution in C++ with detailed explanations (8ms, O(n^2) time and O(n) space)
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