[leetcode] 221. Maximal Square 解题报告

题目链接：https://leetcode.com/problems/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.

思路：一道比较明显的动态规划题目。因为是要寻找一个正方形，所以就比较简单一些，借助一个辅助数组dp[][]，dp[i][j]表示当前正方形的边长，并且其值由两种情况构成：

1. 如果matrix[i-1][j-1] = 1, 因为是正方形，则其值为min(dp[i-1][j-1]，dp[i][j-1]，dp[i-1][j])中的最小值加1，小正方形构成大正方形，状态转移方程即为：

dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]) + 1;

2. 如果matrix[i-1][j-1] = 0，则dp[i][j] = 0;

时间复杂度为O(m*n)，空间复杂度为O(m*n)

代码如下：

class Solution {
public:
int maximalSquare(vector<vector<char>>& matrix) {
if(matrix.size() == 0)
return 0;
int max = 0, dp[matrix.size()+1][matrix[0].size()+1];
memset(dp, 0, sizeof(dp));
for(int i =1; i <= (int)matrix.size(); i++)
{
for(int j =1; j<= (int)matrix[0].size(); j++)
{
if(matrix[i-1][j-1] == '1')
{
dp[i][j] = min(dp[i-1][j-1], min(dp[i-1][j], dp[i][j-1])) + 1;
if(dp[i][j] >= max)
max = dp[i][j];
}
else
dp[i][j] = 0;
}
}
return max*max;
}
};