221. Maximal Square

最新推荐文章于 2021-04-29 20:57:26 发布

BigFatSheep

最新推荐文章于 2021-04-29 20:57:26 发布

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分类专栏： Leetcode Medium

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本文介绍了一种使用动态规划求解二维数组中由1组成的最大正方形面积的问题。通过构建一个二维DP数组，计算每个元素作为正方形右下角的可能性，从而找出最大正方形边长并返回其面积。

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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

Problem URL

Solution

给一个二维数组，数组中有0有1，求由1组成的最大的正方形的面积。在这个数组中正方形肯定是：
1234
2234
3334
4444
这样的形式。所以使用动态规划解决。

We new a two-dimensional array to perform dynamic programming in this problem. If an element in matrix could be a square, then it up, left and up-left should be 1. So we could accord this to calculate the length of the square in the matrix.

The length of this element is the minum value of its up, left and up-left dp.

Code

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