LeetCode 0300 Longest Increasing Subsequence【DP，最长上升子序列LCS】

最新推荐文章于 2022-03-18 14:26:20 发布

原创最新推荐文章于 2022-03-18 14:26:20 发布 · 191 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #dp

LeetCode 专栏收录该内容

51 篇文章

订阅专栏

本文探讨了寻找无序整数数组中最长上升子序列长度的问题，提供了两种算法实现：一种是基于动态规划的O(n^2)复杂度算法，另一种提出了改进至O(nlogn)时间复杂度的可能性。

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

Example:

Input: [10,9,2,5,3,7,101,18]
Output: 4 
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.

Note:

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?

题意

给定一个无序的整数数组，找到其中最长上升子序列的长度。

思路1

dp[i] 表示以当前元素结尾的最长子序列

代码1

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