子序列，不需要連續的。

思路一：傳統的dp，dp[i]表示前i個數且以第i個數字結尾的最長增長子序列，遍歷陣列，dp[i] = max(dp[i], dp[j] + 1)  if nums[i] > nums[j], j from 0 to i

思路二：維護一個數組，vec[i]表示，當遞增序列長度是i的時候最後一個數的最小值

程式碼：

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        // version 1: traditonal dp
        /*
        vector<int> dp(nums.size(), 1);
        int res = 0;
        for (int i = 0; i < nums.size(); ++i) {
            for (int j = 0; j < i; ++j) {
                if (nums[i] > nums[j]) {
                    dp[i] = max(dp[i], dp[j] + 1);
                }
            }
            res = max(res, dp[i]);
        }
        return res;
        */
        // version 2: DP & BS
        /*
        vector<int> dp;
        for (int i = 0; i < nums.size(); ++i) {
            int left = 0, right = dp.size();
            while (left < right) {
                int mid = left + (right - left) / 2;
                if (dp[mid] < nums[i]) left = mid + 1;
                else right = mid;
            }
            if (right >= dp.size()) dp.push_back(nums[i]);
            else dp[right] = nums[i];
        }
        return dp.size();
        */
        // version 3: DP & lower_bound function in C++ STL instead of BS
        vector<int> v;
        for (auto a : nums) {
            auto it = lower_bound(v.begin(), v.end(), a);
            if (it == v.end()) v.push_back(a);
            else *it = a;
        }
        return v.size();
    }
};