221. Maximal Square

难度: Medium

题目描述

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'.

解题思路

代码实现

解决方案

java

func maximalSquare(matrix [][]byte) int {

	dp := make([][]int, len(matrix))
	for i := range dp {
		dp[i] = make([]int, len(matrix[0]))
	}
	result := 0
	for i := 0; i < len(matrix); i++ {
		dp[i][0] = int(matrix[i][0]-'0')
		if dp[i][0] == 1 {
			result = 1
		}

	}
	for i := 0; i < len(matrix[0]); i++ {
		dp[0][i] = int(matrix[0][i]-'0')
		if dp[0][i] == 1 {
			result = 1
		}

	}

	for i := 1; i < len(matrix); i++ {
		for j := 1; j < len(matrix[0]); j++ {
			if matrix[i][j] == '1' {

				up := int(dp[i-1][j])
				left := int(dp[i][j-1])
				leftUp := int(dp[i-1][j-1])

				dp[i][j] = slices.Min([]int{up, left, leftUp}) + 1
				bigger := result < dp[i][j]
				if bigger {
					result = dp[i][j]
				}
			} else {
				dp[i][j] = 0
			}
		}
	}

	return result * result

}