#week12
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?
Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.
分析:
最长不下降子序列
状态转换:
if (nums[j] < nums[i] && (f[j] + 1 > f[i]))
f[i] = f[j] + 1;
题解:
1 class Solution { 2 public: 3 int lengthOfLIS(vector<int>& nums) { 4 int size = nums.size(); 5 if (size == 0) return 0; 6 int f[size], max = 1; 7 f[0] = 1; 8 for (int i = 1; i < size; i++) { 9 f[i] = 1; 10 for (int j = 0; j < i; j++) { 11 if (nums[j] < nums[i] && (f[j] + 1 > f[i])) { 12 f[i] = f[j] + 1; 13 if (f[i] > max) max = f[i]; 14 } 15 } 16 } 17 return max; 18 } 19 };
其他人题解,O(nlogn)
1 class Solution { 2 public: 3 int lengthOfLIS(vector<int>& nums) { 4 vector<int> res; 5 for(int i=0; i<nums.size(); i++) { 6 auto it = std::lower_bound(res.begin(), res.end(), nums[i]); 7 if(it==res.end()) res.push_back(nums[i]); 8 else *it = nums[i]; 9 } 10 return res.size(); 11 } 12 };