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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 }
LeetCode-Longest Increasing Subsequence
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原文地址：http://www.cnblogs.com/lishiblog/p/5724676.html
