[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

 class Solution {

     public int lengthOfLIS(int[] nums) {

         if(nums.length < 2)   return nums.length;

         int result = 1;

         int[] dp = new int[nums.length];

         for(int i = 0; i < nums.length; i++){

             dp[i] = 1;

         }

         for(int i = 1; i< nums.length; i++){

             for(int j = 0; j < i; j++){

                 if(nums[j] < nums[i]){

                     dp[i] = Math.max(dp[i], dp[j]+1);

                 }

             }

             result = Math.max(result, dp[i]);

         }

         return result;

     }

 }

[leetcode]300. Longest Increasing Subsequence最长递增子序列的相关教程结束。