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?

 1 public class Solution {
 2     public int lengthOfLIS(int[] nums) {
 3         int[] dp = new int[nums.length];
 4         int res = 0;
 5         for(int i = 0;i<nums.length;i++){
 6             dp[i] = 1;
 7             for(int j=0;j<i;j++){
 8                 if(nums[i]>nums[j]){
 9                     dp[i] = Math.max(dp[j]+1,dp[i]);
10                 }
11             }
12             res = Math.max(res,dp[i]);
13         }
14         return res;
15     }
16 }
17 //the run time complexity could be O(n^2), the space complexity could be O(n);

第二种方法比较不好理解，当我们遍历数组中的每一个元素的时候，我们可以将该元素在dp中进行排序，如果该元素大于dp中的最后一个元素，则长度+1，而如果该元素不比所有元素都打，则排序到所对应的位置，更新新元素，代码如下：

 1 public class Solution {
 2     public int lengthOfLIS(int[] nums) {
 3         int[] dp = new int[nums.length];
 4         int len = 0;
 5         for(int num:nums){
 6             int i=binarysearch(dp,0,len,num);
 7             if(i==len) len++;
 8         }
 9         return len;
10     }
11     public int binarysearch(int[] dp,int left,int right,int target){
12         while(left<right){
13             int mid = left+(right-left)/2;
14             if(dp[mid]>=target) right = mid;
15             else if(dp[mid]<target) left = mid+1;
16         }
17         dp[left] = target;
18         return left;
19     }
20 }
21 //the time complexity could be O(nlogn),the space complexity could be O(n)