题目:原题链接(中等)

标签:动态规划、数组

解法 时间复杂度 空间复杂度 执行用时
Ans 1 (Python) O ( M × N ) O(M×N) O(M×N) O ( M × N ) O(M×N) O(M×N) 160ms (88.15%)
Ans 2 (Python)
Ans 3 (Python)

解法一:

class Solution:
    def countSquares(self, matrix: List[List[int]]) -> int:
        m, n = len(matrix), len(matrix[0])

        dp = [[0] * n for _ in range(m)]

        ans = 0

        for i in range(m):
            for j in range(n):
                if matrix[i][j] == 1:
                    if i == 0 or j == 0:
                        dp[i][j] = 1
                        ans += 1
                    else:
                        dp[i][j] = min(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]) + 1
                        ans += dp[i][j]

        return ans