[LeetCode] 221. Maximal Square

Given an m x n binary matrix filled with 0’s and 1’s, find the largest square containing only 1’s and return its area.

Example 1:

Input: matrix = [[“1”,”0”,”1”,”0”,”0”],[“1”,”0”,”1”,”1”,”1”],[“1”,”1”,”1”,”1”,”1”],[“1”,”0”,”0”,”1”,”0”]]
Output: 4

Example 2:

Input: matrix = [[“0”,”1”],[“1”,”0”]]
Output: 1

Example 3:
Input: matrix = [[“0”]]
Output: 0

Constraints:
m == matrix.length
n == matrix[i].length
1 <= m, n <= 300
matrix[i][j] is ‘0’ or ‘1’.

最大正方形。

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。

思路

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。这是一道矩阵类的DP问题。dp[i][j] 的定义是以坐标 [i][j] 为右下角的点组成的正方形的最大边长。状态转移方程是dp(i, j) = min(dp(i−1, j), dp(i−1, j−1), dp(i, j−1)) + 1。这样的题目练多了才会有思路。

复杂度

时间O(n^2)
空间O(n^2)

代码

Java一维实现

class Solution {
    public int maximalSquare(char[][] matrix) {
        int m = matrix.length;
        int n = m > 0 ? matrix[0].length : 0;
        int[] dp = new int[n + 1];
        int max = 0;
        int prev = 0;
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                int temp = dp[j];
                if (matrix[i - 1][j - 1] == '1') {
                    dp[j] = Math.min(Math.min(dp[j - 1], prev), dp[j]) + 1;
                    max = Math.max(max, dp[j]);
                } else {
                    dp[j] = 0;
                }
                prev = temp;
            }
        }
        return max * max;
    }
}

Java二维实现

class Solution {
    public int maximalSquare(char[][] matrix) {
        int m = matrix.length;
        int n = matrix[0].length;
        int[][] dp = new int[m + 1][n + 1];
        int res = 0;
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (matrix[i - 1][j - 1] == '1') {
                    dp[i][j] = min(
                        dp[i - 1][j] + 1,
                        dp[i][j - 1] + 1,
                        dp[i - 1][j - 1] + 1
                    );
                    res = Math.max(res, dp[i][j]);
                }
            }
        }
        return res * res;
    }

    private int min(int a, int b, int c) {
        return Math.min(a, Math.min(b, c));
    }
}

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