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?
Analyse: For every number in nums, check from index 0 to it former number, if it is larger than a number, update the array.
Runtime: 124ms.
1 class Solution { 2 public: 3 int lengthOfLIS(vector<int>& nums) { 4 int n = nums.size(); 5 if(n < 2) return n; 6 7 vector<int> dp(n, 1); 8 dp[0] = 1; 9 int result = 0; 10 for(int i = 1; i < nums.size(); i++) { 11 for(int j = 0; j < i; j++) { 12 if(nums[i] > nums[j]) 13 dp[i] = max(dp[i], dp[j] + 1); 14 } 15 result = max(result, dp[i]); 16 } 17 return result; 18 } 19 };