[leetcode]300. Longest Increasing Subsequence最长递增子序列

Given an unsorted array of integers, find the length of longest increasing subsequence.

Example:

Input: [10,9,2,5,3,7,101,18]
Output: 4 
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4. 

Note:

  • 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?

Solution: DP

dp[i] stands for the length of LIS ends with nums[i] (nums[i] must be used)

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code

 1 class Solution {
 2     public int lengthOfLIS(int[] nums) {
 3         if(nums.length < 2)   return nums.length;
 4         
 5         int result = 1;
 6         int[] dp = new int[nums.length];
 7       
 8         for(int i = 0; i < nums.length; i++){
 9             dp[i] = 1;
10         }
11         
12         for(int i = 1; i< nums.length; i++){
13             for(int j = 0; j < i; j++){
14                 if(nums[j] < nums[i]){
15                     dp[i] = Math.max(dp[i], dp[j]+1);
16                 } 
17             }
18             result = Math.max(result, dp[i]);     
19         }
20         return result;
21     }
22 }
原文地址:https://www.cnblogs.com/liuliu5151/p/9841060.html