Leetcode 300. Longest Increasing Subsequence-优快云博客

本文介绍了一种寻找整数数组中最长递增子序列长度的算法，包括O(n^2)复杂度的经典动态规划解法及更高效的O(nlogn)复杂度解法，并附带示例代码。

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问题描述

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?

Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.

O(n2)解法：
定义数组dp[i]表示前i个数的最长递增序列的长度，则状态转移矩阵：
dp[i]=max(dp[j])+1,(j

    public int lengthOfLIS(int[] nums) {
        if(nums==null)
            return 0;
        int n=nums.length;
        if(n==0)
            return 0;
        int[]dp=new int[n];
        dp[0]=1;
        int ans=1;
        for(int i=1;i<n;i++){
            for(int j=0;j<i;j++){
                if(nums[i]>nums[j]&&dp[i]<dp[j])
                    dp[i]=dp[j];
            }
            dp[i]++;
            if(ans<dp[i])
                ans=dp[i];
        }
        return ans;

    }

O(nlogn)解法：二分查找
维护一个递增的序列，对其进行二分查找当前元素nums[i]，覆盖原有元素或者在末尾添加。
代码：

    public int lengthOfLIS(int[] nums) {
        if(nums==null)
            return 0;
        int n=nums.length;
        if(n==0)
            return 0;
        int[]dp=new int[n];
        dp[0]=nums[0];
        int len=1;
        for(int i=1;i<n;i++){
            int low=0,high=len-1;
            while(low<=high){
                int mid=low+(high-low)/2;
                if(nums[i]<=dp[mid])
                    high=mid-1;
                else
                    low=mid+1;
            }
            if(low>=len)
                dp[len++]=nums[i];
            else
                dp[low]=nums[i];
        }
        return len;

    }