[leetcode] 300. Longest Increasing Subsequence

最新推荐文章于 2025-08-10 20:50:18 发布

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本文探讨了如何在未排序整数数组中找到最长递增子序列长度的问题，并提供了一种运行时间为O(n^2)的解决方案。同时，文章还讨论了如何进一步优化算法至O(n log n)的时间复杂度。

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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?

解法一：

借鉴了之前increasing triplet sequence的思路。

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        if(nums.size()<1) return 0;
        vector<int> seq(1,INT_MAX);
        
        for(int i=0; i< nums.size(); i++){
            for(int j=0; j<seq.size(); j++){
                if(nums[i]<=seq[j]){
                    seq.erase(seq.begin()+j);
                    seq.insert(seq.begin()+j,nums[i]);
                    break;
                }
            }
            if(nums[i]>seq.back()) seq.push_back(nums[i]);
        }
        return seq.size();
    }
};