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?
Analysis:
https://segmentfault.com/a/1190000003819886
tails[i]: for an increasing subsequence with length i, what is the tail element in nums[].
Solution:
1 public class Solution { 2 public int lengthOfLIS(int[] nums) { 3 if (nums.length==0) return 0; 4 5 int[] tails = new int[nums.length]; 6 tails[0] = 0; 7 8 int res = 0; 9 for (int i=1;i<nums.length;i++){ 10 int pos = findPos(nums,tails,nums[i],res); 11 if (pos!=-1) tails[pos] = i; 12 13 if (pos>res) res = pos; 14 } 15 return res+1; 16 } 17 18 public int findPos(int[] nums, int[] tails, int val, int end){ 19 int start = 0; 20 while (start<=end){ 21 int mid = start + (end-start)/2; 22 if (val == nums[tails[mid]]) return -1; 23 else if (val < nums[tails[mid]]){ 24 end = mid-1; 25 } else { 26 start = mid+1; 27 } 28 } 29 return start; 30 } 31 }