leetcode

题目描述：

 

 方法一：动态规划 O(mn) O(mn)

class Solution:
    def findLength(self, A: List[int], B: List[int]) -> int:
        n, m = len(A), len(B)
        dp = [[0] * (m + 1) for _ in range(n + 1)]
        ans = 0
        for i in range(n - 1, -1, -1):
            for j in range(m - 1, -1, -1):
                dp[i][j] = dp[i + 1][j + 1] + 1 if A[i] == B[j] else 0
                ans = max(ans, dp[i][j])
        return ans

语言	方法
2370	r0sMJ
XV1wH	摆地摊一个月真实故事「血泪史」提醒各位
3624	2009.01.05 09-57-35

方法二：滑动窗口 O((m+n) * min(m,n)) O(1)

class Solution:
    def findLength(self, A: List[int], B: List[int]) -> int:
        def maxLength(addA: int, addB: int, length: int) -> int:
            ret = k = 0
            for i in range(length):
                if A[addA + i] == B[addB + i]:
                    k += 1
                    ret = max(ret, k)
                else:
                    k = 0
            return ret
        
        n, m = len(A), len(B)
        ret = 0
        for i in range(n):
            length = min(m, n - i)
            ret = max(ret, maxLength(i, 0, length))
        for i in range(m):
            length = min(n, m - i)
            ret = max(ret, maxLength(0, i, length))
        return ret

方法三：二分+哈希表  留坑*