Longest Increasing Subsequence

300. Longest Increasing Subsequence Given an integer array  ums`, return the length of the longest strictly increasing subsequence.

Given a[] f[i] The length of the longest increasing subsequence with the last element being a[i]

1,&\forall 1\le j \le i, a_j < a_i\\ \max\limits_{1\le j \le i, a_j < a_i}​(f[j])+1,​&otherwise.​ \end{cases}

Answer: $\underset{1\le i\le n}{\max }(f[i])$

int lengthOfLIS(vector<int> &nums) {
    vector<int> f(nums.size());
    int n = nums.size();
    for (int i = 0; i < n; i++) {
      f[i] = 1;
      int maxv = 0;
      for (int j = 0; j < i; j++) {
        if (nums[i] > nums[j]) {
          maxv = max(maxv, f[j]);
        }
      }
      f[i] = maxv + 1;
    }
    return *max_element(f.begin(), f.end());
  }