采用的方法是动态规划，然后用一个list保存最长子路径，最后逆序输出这个list就是最长子序列。

class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        dp = [1 for _ in range(len(nums))]
        dp[0] = 1
        route = [i for i in range(len(nums))]
        maxlen = 0
        k = 0
        for i in range(1, len(nums)):
            for j in range(i):
                if nums[i]> nums[j]:
                    if dp[i] < dp[j]+1:
                        dp[i] = dp[j]+1
                        if maxlen < dp[i]:
                            maxlen = dp[i]
                            k = i
                        route[i] = j #保存路径
        res = [i for i in range(maxlen)]
        print(route)
        j = maxlen-1
        res[j] = nums[k]
        j -= 1
        o = k
        while route[o]!=o:#逆序打印route即为子序列
            
            o = route[o]
            #print(j, o)
            res[j] = nums[o]
            j -= 1
        print(res)
        return maxlen