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Given a sequence of integers, find the longest increasing subsequence (LIS).
You code should return the length of the LIS.
For [5, 4, 1, 2, 3], the LIS is [1, 2, 3], return 3
For [4, 2, 4, 5, 3, 7], the LIS is [4, 4, 5, 7], return 4
Time complexity O(n^2) or O(nlogn)
What‘s the definition of longest increasing subsequence?
* The longest increasing subsequence problem is to find a subsequence of a given sequence in which the subsequence‘s elements are in sorted order, lowest to highest, and in which the subsequence is as long as possible. This subsequence is not necessarily contiguous, or unique.
* https://en.wikipedia.org/wiki/Longest_common_subsequence_problem
Solution 1 (nlogn):
1 public class Solution { 2 /** 3 * @param nums: The integer array 4 * @return: The length of LIS (longest increasing subsequence) 5 */ 6 public int longestIncreasingSubsequence(int[] nums) { 7 if (nums.length==0) return 0; 8 int len = nums.length; 9 int[] seqEnd = new int[len+1]; 10 seqEnd[1] = 0; 11 int lisLen = 1; 12 for (int i=1;i<len;i++){ 13 int pos = findPos(nums,seqEnd,lisLen,i); 14 seqEnd[pos] = i; 15 if (pos>lisLen) lisLen = pos; 16 } 17 18 return lisLen; 19 20 } 21 22 public int findPos(int[] nums, int[] seqEnd, int lisLen, int index){ 23 int start = 1; 24 int end = lisLen; 25 while (start<=end){ 26 int mid = (start+end)/2; 27 28 if (nums[index] == nums[seqEnd[mid]]){ 29 return mid; 30 } else if (nums[index]>nums[seqEnd[mid]]){ 31 start = mid+1; 32 } else end = mid-1; 33 } 34 return start; 35 } 36 }
LintCode-Longest Increasing Subsequence
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原文地址：http://www.cnblogs.com/lishiblog/p/4190936.html
