码迷,mamicode.com
首页 > 其他好文 > 详细

300. Longest Increasing Subsequence

时间:2017-04-04 20:53:02      阅读:164      评论:0      收藏:0      [点我收藏+]

标签:i++   logs   example   sequence   color   sort   ref   algorithm   ==   

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?

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        vector<int> res;
        for(int i=0; i<nums.size(); i++) {
            auto it = std::lower_bound(res.begin(), res.end(), nums[i]);
            if(it==res.end()) res.push_back(nums[i]);
            else *it = nums[i];
        }
        return res.size();
    }
};

 

300. Longest Increasing Subsequence

标签:i++   logs   example   sequence   color   sort   ref   algorithm   ==   

原文地址:http://www.cnblogs.com/tsunami-lj/p/6665872.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!