LeetCode-221. Maximal Square

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Description

Given a 2D binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.

Example

Input: 

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Output: 4

Solution 1(C++)

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

算法分析

这道题最关键的就是问题的转化，题目要求最大方阵的面积，直接来求还是蛮麻烦的。但是考虑到方阵的计算特性，我们可以转换为找到最大方阵的边长。

首先要注意，矩阵的元素是char类型，所以转化为int类型：matrix[i][j] - ‘0’；

构建动态规划数组，size[i][j]表示matrix[i][j]的最大边长。状态转移方程如下：

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

程序分析

略。