[LeetCode] 221. Maximal Square（最大正方形）

• Difficulty: Medium

• Related Topics: Dynamic Programming

• Link: https://leetcode.com/problems/maximal-square/

Description

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.
给定一个 m x n 的二进制矩阵 matrix，其中每个元素为 0 或 1，找到只包含 1 的面积最大的正方形并返回它的面积。

Examples

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'.

Solution

如果 matrix[i][j] 能组成正方形的一部分，则其左上方的3 个元素也应该是正方形的一部分。

使用动态规划的方法，记 size[i][j] 表示以 matrix[i][j] 作为右下角的正方形的最大边长，初始时，size[i][j] 均为 0。则：

1. 只有当 matrix[i][j] 为 1 的时候，才更新 size

2. size[i][j] = min(size[i - 1][j], size[i][j - 1], size[i - 1][j - 1]) + 1

第 2 步的 +1 表示如下意思：如果 matrix[i][j] 能够与其他元素构成一个更大的正方形，那么这个正方形的边长会 +1，否则，其本身就是一个 1 x 1 的正方形。

代码如下：

import kotlin.math.max

class Solution {
    fun maximalSquare(matrix: Array<CharArray>): Int {
        if (matrix.isEmpty() || matrix[0].isEmpty()) {
            return 0
        }
        val size = Array(matrix.size) { IntArray(matrix[0].size) }
        var maxSize = 0

        for (i in matrix.indices) {
            size[i][0] = if (matrix[i][0] == '1') 1 else 0
            maxSize = max(maxSize, size[i][0])
        }
        for (i in matrix[0].indices) {
            size[0][i] = if (matrix[0][i] == '1') 1 else 0
            maxSize = max(maxSize, size[0][i])
        }

        for (i in 1..matrix.lastIndex) {
            for (j in 1..matrix[i].lastIndex) {
                if (matrix[i][j] == '1') {
                    size[i][j] = minOf(size[i - 1][j], size[i][j - 1], size[i - 1][j - 1]) + 1
                    maxSize = max(maxSize, size[i][j])
                }
            }
        }

        return maxSize * maxSize
    }
}