300. Longest Increasing Subsequence

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本文探讨了如何在未排序整数数组中寻找最长递增子序列长度的问题，并提供了一个运行时间为O(n²)的解决方案。同时，文章还讨论了进一步优化到O(n log n)的可能性。

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题目：

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?

解题思路：

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        int n = nums.size();
        if(n==0)return 0;
        vector<int> res(n,1);
        for(int cur=1;cur<n;cur++)
            for(int pre=0;pre<cur;pre++)
                if(nums[pre]<nums[cur])res[cur]=max(res[cur],res[pre]+1);
        return *max_element(res.begin(),res.end());
    }
};