欢迎访问LeetCode题解合集

题目描述

在一个由 '0' 和 '1' 组成的二维矩阵内，找到只包含 '1' 的最大正方形，并返回其面积。

示例 1：

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输入：matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
输出：4

示例 2：

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输入：matrix = [["0","1"],["1","0"]]
输出：1

示例 3：

输入：matrix = [["0"]]
输出：0

提示：

• m == matrix.length

• n == matrix[i].length

• 1 <= m, n <= 300

• matrix[i][j] 为 '0' 或 '1'

题解：

动态规划。

设 f[i][j] 表示以 (i,j) 为正方形右下角，只包含 1 的正方形的变长最大值。

转移方程：

• 如果 i == 0 || j == 0 ，f[i][j] = matrix[i][j] - '0'
• 如果 matrix[i][j] == '0' ，f[i][j] = 0
• 否则的话， f[i][j] = min{f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]} + 1

时间复杂度：$O(n^2)$

额外空间复杂度：$O(n^2)$

class Solution {
public:
    int maximalSquare(vector<vector<char>>& matrix) {
        int n = matrix.size();
        int m = matrix[0].size();
        vector<vector<int>> f(n, vector<int>(m));
        int ret = 0;
        for ( int i = 0; i < n; ++i ) {
            for ( int j = 0; j < m; ++j ) {
                if ( !i || !j )
                    f[i][j] = matrix[i][j] & 15;
                else if ( matrix[i][j] == '0' )
                    f[i][j] = 0;
                else
                    f[i][j] = min( {f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]} ) + 1;
                ret = max( ret, f[i][j] );
            }
        }
        return ret * ret;
    }
};
/*
时间：28ms，击败：75.32%
内存：11.2MB，击败：74.87%
*/

使用滚动数组优化：

class Solution {
public:
    int maximalSquare(vector<vector<char>>& matrix) {
        int n = matrix.size();
        int m = matrix[0].size();
        vector<int> f(m);
        int ret = 0;
        for ( int i = 0; i < m; ++i ) {
            f[i] = matrix[0][i] & 15;
            ret |= f[i];
        }
        int pre, tmp;
        for ( int i = 1; i < n; ++i ) {
            pre = f[0];
            f[0] = matrix[i][0] & 15;
            if ( f[0] && !ret ) ret = 1;
            for ( int j = 1; j < m; ++j ) {
                tmp = f[j];
                if ( matrix[i][j] == '0' )
                    f[j] = 0;
                else
                    f[j] = min( {f[j], f[j - 1], pre} ) + 1;
                pre = tmp;
                ret = max( f[j], ret );
            }
        }
        return ret * ret;
    }
};
/*
时间：24ms，击败：91.69%
内存：10.5MB，击败：98.54%
*/