# Dynamic programming.
class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        if not nums: return 0
        dp = [1] * len(nums)
        for i in range(len(nums)):
            for j in range(i):
                if nums[j] < nums[i]: # 如果要求非严格递增，将此行 '<' 改为 '<=' 即可。
                    dp[i] = max(dp[i], dp[j] + 1)
        return max(dp)

class Solution:
    def findLengthOfLCIS(self, nums: List[int]) -> int:
        l = 0  # 起点
        r = 0  # 终点
        maxLen = 0  # 待返回的最长长度

        while r < len(nums):
            if r == l or nums[r - 1] < nums[r]:  # 保持递增
                maxLen = max(maxLen, r - l + 1)  # 比较取大
                r += 1  # 终点前进
            else:  # 递增中断
                l = r  # 更新起点

        return maxLen

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