DP问题：leetcode(300) Longest Increasing Subsequence

问题描述：

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?

主要思路：使用了时间复杂度为O(N^2)的动态规划。设a[i]为以nums[i]为最后一个数所构成的最长递增子序列的长度，则初始maxn=-1,j从0遍历到i-1，如果nums[i]>nums[j],则maxn=max{maxn,a[j+1]}，否则maxn=max{ maxn,1 },循环一次后a[i]=maxn。最后找到a数组中的最大值即为所求长度。

代码：

#include<iostream>
#include<vector>
#include<cstring>
#include<cmath>
using namespace std;

class Solution {
public:
int lengthOfLIS(vector<int>& nums) {
int a[4000],len=nums.size(),maxn,maxr=1;
a[0]=1;
if(len==0) return 0;
for(int i=1;i<len;i++){
maxn=-1;
for(int j=0;j<=i-1;j++){
if(nums[i]>nums[j]) maxn=maxn>a[j]+1?maxn:a[j]+1;
else maxn=maxn>1?maxn:1;
}
a[i]=maxn;
maxr=maxr>maxn?maxr:maxn;
}
return maxr;
}

};

int main()
{
//int arr[]={10,9,2,5,3,7,101,18};
int arr[]={1,3,6,7,9,4,10,5,6};
vector<int>nums(arr,arr+sizeof(arr)/sizeof(int));
cout<<(new Solution())->lengthOfLIS(nums);
}