Longest Increasing Subsequence

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.

解决方法：利用动态规划来做，设dp[ i ]为以nums[ i ]为尾的LIS的长度，初始化dp[ i ]=0.

则 dp[ i ]=max{ dp[ j ] | nums[ j ]<nums[ i ], 0<=j<i }+1.

复杂度为O(n*n).

具体代码为：

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