LeetCode:Maximal Square

来源：互联网发布：淘宝模特多少钱一个月编辑：程序博客网时间：2024/06/13 19:49

Maximal Square

Total Accepted: 25468 Total Submissions: 110270 Difficulty: Medium

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 01 0 1 1 11 1 1 1 11 0 0 1 0

Return 4.

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

Subscribe to see which companies asked this question

Hide Tags

Dynamic Programming

Hide Similar Problems

(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<vector<char>>& matrix) {                int m = matrix.size();        if(m==0) return 0;        int n = matrix[0].size();        vector<vector<int>> ret(m, vector<int>(n,0));        int maxSize = 0;                for(int i=0;i<m;i++) {            ret[i][0] = matrix[i][0]-'0';            maxSize = max(ret[i][0], maxSize);        }        for(int j=0;j<n;j++) {            ret[0][j] = matrix[0][j]-'0';            maxSize = max(ret[0][j], maxSize);        }        for(int i=1;i<m;i++)        for(int j=1;j<n;j++) {            if(matrix[i][j]=='1') ret[i][j] = min(ret[i-1][j],min(ret[i][j-1],ret[i-1][j-1])) + 1;            maxSize = max(ret[i][j], maxSize);        }        return maxSize*maxSize;    }};

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)

0 0