面试经典150题 P221 最大正方形

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时间轴

时间轴

2025-12-15

init

动态规划

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题目：

P221 最大正方形

https://leetcode.cn/problems/maximal-square/description/?envType=study-plan-v2&envId=top-interview-150

dp[i][j]暴力更新状态

#include <vector>
using std::vector;

class Solution {
    public:
        int maximalSquare(vector<vector<char> > &matrix)
        {
                int i, j, k, m = matrix.size(), n = matrix[0].size();
                // dp[i][j]表示以matrix[i][j]为右下角的全1正方形大小
                vector<vector<int> > dp(m, vector<int>(n, 0));
                int max_len = 0;
                for (i = 0; i < m; i++) {
                        for (j = 0; j < n; j++) {
                                if (matrix[i][j] == '1') {
                                        dp[i][j] = 1;
                                        max_len = 1;
                                }
                        }
                }

                bool only1;
                for (i = 1; i < m; i++) {
                        for (j = 1; j < n; j++) {
                                if (dp[i][j] == 1 && dp[i - 1][j - 1] != 0) {
                                        // dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1
                                        only1 = true;
                                        for (k = 1; k <= dp[i - 1][j - 1]; k++) { //boundary
                                                if (dp[i][j - k] == 0) {
                                                        only1 = false;
                                                        break;
                                                }
                                                if (dp[i - k][j] == 0) {
                                                        only1 = false;
                                                        break;
                                                }
                                        }
                                        // k-1
                                        if (only1) {
                                                dp[i][j] = dp[i - 1][j - 1] + 1;
                                        } else {
                                                dp[i][j] = (k - 1) + 1;
                                        }
                                        max_len = std::max(dp[i][j], max_len);
                                }
                        }
                }
                return max_len * max_len;
        }
};

实际 dp 更新方程:
when dp[i][j] == 1:

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

#include <vector>
#include <algorithm>
using std::vector;

class Solution {
    public:
        int maximalSquare(vector<vector<char> > &matrix)
        {
                int i, j, m = matrix.size(), n = matrix[0].size();
                // dp[i][j]表示以matrix[i][j]为右下角的全1正方形大小
                vector<vector<int> > dp(m, vector<int>(n, 0));
                int max_len = 0;
                for (i = 0; i < m; i++) {
                        for (j = 0; j < n; j++) {
                                if (matrix[i][j] == '1') {
                                        dp[i][j] = 1;
                                        max_len = 1;
                                }
                        }
                }

                bool only1;
                for (i = 1; i < m; i++) {
                        for (j = 1; j < n; j++) {
                                if (dp[i][j] == 1) {
                                        dp[i][j] = std::min({ dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1] }) + 1;
                                        max_len = std::max(dp[i][j], max_len);
                                }
                        }
                }
                return max_len * max_len;
        }
};