LeetCode 300. Longest Increasing Subsequence 题解

最新推荐文章于 2024-02-20 16:01:51 发布

原创最新推荐文章于 2024-02-20 16:01:51 发布 · 319 阅读

0 ·

CC 4.0 BY-SA版权

leetcode 专栏收录该内容

38 篇文章

订阅专栏

本文探讨了如何求解给定整数数组中最长上升子序列的长度问题，并提供了一种时间复杂度为O(n²)的解决方案。通过动态规划的方法，文章详细解释了如何计算每个元素作为子序列结尾时可能的最长子序列长度。

300. Longest Increasing Subsequence

My Submissions

Total Accepted: 55308
Total Submissions: 149684
Difficulty: Medium
Contributors: Admin

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：

令数组为a[n]。

设以数组中第i个数为结尾的最长上升子序列的长度为d[i]。则d[i]=max(d[j]+1),(0<=j<i&&a[j]<a[i])。

代码展示：

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        vector<int> ans;
        int n=nums.size();
        if(!n) return 0;
        ans.push_back(1);
        int ans_max=1;
        for(int i=1;i<n;i++)
        {
            int tmp_max=0;
            for(int j=0;j<i;j++)
            {
                if(nums[j]<nums[i])
                {
                    tmp_max=max(tmp_max,ans[j]);
                }
            }
            ans.push_back(tmp_max+1);
            ans_max=max(ans_max,tmp_max+1);
        }
        return ans_max;
    }
};