LeetCode OJ：Longest Increasing Subsequence（最长递增序列）

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?

dp解决，注意这里的递增序列不是指连续的递增，可以是不连续的，代码如下：

 class Solution {

 public:

     int lengthOfLIS(vector<int>& nums) {

         if(nums.size() <= ) return nums.size();

         vector<int> dp(nums.size(), );

         int maxVal = ;

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

             dp[i] = ;

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

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

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

                     maxVal = max(dp[i], maxVal);

                 }

             }

         }

         return maxVal;

     }

 };

巴特西

LeetCode OJ：Longest Increasing Subsequence（最长递增序列）

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