221. Maximal Square

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

题目是让求给定的矩阵中由1构成的正方形的最大面积，找到求一个点(i,j)的公式，如果matrix[i][j] == '0',那么以(i, j)为断点的正方形边长为0，如果matrix[i][j] == '1';

详细分析

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;
    }
};

压缩上面的矩阵 c++