LeetCode 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 is4. 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：从前向后求出每个元素的递增序列个数，而当前 i 的递增序列个数需要扫描 0~i-1 的元素，找到比 i 元素的值小的元素，且递增序列数最大的数，再加1，就是当前 i 的递增序列个数。

public class Solution {    public int lengthOfLIS(int[] nums) {        if(nums.length<=0){            return 0;        }else{            int []dp= new int[nums.length];            dp[0]=1;            int max=0;            for(int i=0;i<nums.length;i++){                max=0;                for(int j=0;j<i;j++){                    if(nums[j]<nums[i]){                        max=max>dp[j]?max:dp[j];                    }                }                dp[i]=max+1;            }            for(int i=0;i<dp.length;i++){                max=max>=dp[i]?max:dp[i];            }            return max;        }    }}