300. Longest Increasing Subsequence

Problem statement:

Given an unsorted array of integers, find the length of longest increasing subsequence.

For example,
Given [10, 9, 2, 5, 3, 7, 101, 18],
The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4. Note that there may be more than one LIS combination, it is only necessary for you to return the length.

Your algorithm should run in O(n2) complexity.

Follow up: Could you improve it to O(n log n) time complexity?

Solution:

This is one sequence DP problem. The dp array is one dimension. dp[i] means first i chars in the given string. The return value is not dp[n], it is one max value among dp[0 ... n - 1].

dp[i] = max(dp[i], dp[j] + 1) if nums[i] == nums[j], meanwhile, update the max LIS.

Time complexity is O(n * n).

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        int max_lis = 0;
        int size = nums.size();
        vector<int> lis(size, 1);
        for(int i = 0; i < size; i++){
            for(int j = 0; j < i; j++){
                if(nums[i] > nums[j]){
                    lis[i] = max(lis[i], lis[j] + 1);
                }
            }
            max_lis = max(max_lis, lis[i]);
        }
        return max_lis;
    }
};