1 class Solution:
 2     def lengthOfLIS(self, nums: List[int]) -> int:
 3       Len = len(nums)
 4       if Len == 0:
 5         return 0
 6       dp = [1] * Len  # 子序列的最短长度是1
 7       for i in range(Len):
 8         for j in range(i):
 9           if nums[j] < nums[i]:
10             dp[i] = max(dp[i], dp[j]+1)
11       maxLen = 1
12       for i in range(Len):
13          if dp[i] > maxLen:
14            maxLen = dp[i]
15       return maxLen

 1 class Solution {
 2 public:
 3     int lengthOfLIS(vector<int>& nums) {
 4       int Len = nums.size();
 5       if(Len == 0)
 6         return 0;
 7       vector<int> dp(Len, 1);  // 子序列的最短长度是1
 8       for(int i = 0; i < Len; i++) {
 9         for(int j = 0; j < i; j++)
10           if(nums[j] < nums[i])
11             dp[i] = max(dp[j]+1, dp[i]);
12       }
13       int maxLen = 1;
14       for(int i = 0; i < Len; i++)
15         if(dp[i] > maxLen)
16           maxLen = dp[i];
17       return maxLen;
18     }
19 };