Leetcode 300. Longest Increasing Subsequence

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;    }