Longest Increasing Subsequence
Total Accepted: 9797 Total Submissions: 30801 Difficulty: Medium
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?
1.o(n*n)
class Solution { public: int lengthOfLIS(vector<int>& nums) { int n = nums.size(); int res = n==0 ? 0 : 1; vector<int> help(n,1); for(int i=1;i<n;i++){ for(int j=0;j<i;j++){ if(nums[j] < nums[i]){ help[i] = max(help[i],help[j]+1); res = max(res,help[i]); } } } return res; } };
2.o(n*lgn)
class Solution { public: int lengthOfLIS(vector<int>& nums) { int n = nums.size(); vector<int> help; for(int i=0;i<n;i++){ auto iter = lower_bound(help.begin(),help.end(),nums[i]); if(iter == help.end()){ help.push_back(nums[i]); }else{ *iter=nums[i]; } } return help.size(); } };
写者:zengzy
出处: http://www.cnblogs.com/zengzy
标题有【转】字样的文章从别的地方转过来的,否则为个人学习笔记
 
                    
                     
                    
                 
                    
                 
                
            
         
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浙公网安备 33010602011771号