300. Longest Increasing Subsequence**

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?

My code:

public class Solution {    public int lengthOfLIS(int[] nums) {        int n=nums.length;        int maxresult=1;        if(n==0) return 0;        int result[]=new int[n];        Arrays.fill(result,1);        for(int i=1;i<n;i++){            for(int j=0;j<i;j++){                if(nums[i]>nums[j]){                    result[i]=Math.max(result[j]+1,result[i]);                }            }            maxresult=Math.max(maxresult,result[i]);        }        return maxresult;    }}

Reference

public class Solution {    public int lengthOfLIS(int[] nums) {                    int[] dp = new int[nums.length];        int len = 0;        for(int x : nums) {            int i = Arrays.binarySearch(dp, 0, len, x);            if(i < 0) i = -(i + 1);            dp[i] = x;            if(i == len) len++;        }        return len;    }}

dp中存储的永远都是nums[0:len]中的递增子序列中最小的元素。

总结：自己的方法是O(n^2)