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?

Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.

思路：

经典dp问题，定义dp数组，dp[i]代表0到i区间的最长递增子序列。状态转移方程为

dp[i] = max(dp[i], dp[j] + 1)。

int lengthOfLIS(vector<int>& nums)
     {
         int n = nums.size(),ret = 0;
         if (n == 0) return 0;
         vector<int>dp(n, 1);
         for (int i = 0; i < n;i++)
         {
             for (int j = 0; j < i; j++)
             {
                 if(nums[i]>nums[j])dp[i] = max(dp[i], dp[j] + 1);                 
             }
             ret = max(ret, dp[i]);
         }
         return ret;         
     }

参考：

http://www.cnblogs.com/yrbbest/p/5047816.html

[leetcode-300-Longest Increasing Subsequence]