300. Longest Increasing Subsequence 类别：动态规划 难度：medium

题目：

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.

思路：

对于每一个i，对于j < i，如果nums[j] < nums[i]，则转移方程为dp[i] = max(dp[i],dp[j] + 1)。

程序：

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