首页 > 代码库 > 动态规划系列【2】最长递增子序列LIS
动态规划系列【2】最长递增子序列LIS
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. Subscribe to see which companies asked this question.
题意:无需多说。主要是这玩意有基础版O(n^2)和增强版O(nlogn)
public class Solution { public int lengthOfLIS(int[] nums) { int length=nums.length; if(length==0)return 0; int max=1; int[] d=new int[length]; for(int i=1;i<length;i++){ d[i]=1; for(int j=0;j<i;j++){ if(nums[j]<nums[i]) d[i]=Math.min(d[i],d[j]=1); } max=Math.max(d[i],max); } return max; } }
PS:思路参加http://www.hawstein.com/posts/dp-novice-to-advanced.html
public class AscentSequence { public int findLongest(int[] A, int n) { // write code here int[] dp=new int[n]; int[] ends=new int[n]; ends[0]=A[0]; dp[0]=1; int max=1; int r=0,right=0,l=0,m=0; for(int i=1;i<n;i++){ l=0; r=right; while(l<=r){ m=(l+r)/2; if(A[i]>ends[m]){ l=m+1; }else{ r=m-1; } } //没明白 right=Math.max(l,right); ends[l]=A[i]; dp[i]=l+1; max=Math.max(dp[i],max); } return max; } }
用二分查找、辅助数组。
动态规划系列【2】最长递增子序列LIS
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。