标签：null pre mat log other new dynamic lis imu
Given an array of positive number, find maximum sum subsequence such that elements in this subsequence are not adjacent to each other.
Recursive formula: f(n) = Math.max{f(n - 1), f(n - 2) + arr[n - 1]}.
Dynamic programming is used to get rid of the overlapping subproblems.
State: T[i]: the max sum of non-adjacent subsequence from arr[0.... n - 1];
Function: T[i] = Math.max(T[i - 1], T[i - 2] + arr[i - 1]);
Init: T[0] = 0, T[1] = arr[0];
Answer: T[arr.length].
1 import java.util.ArrayList; 2 3 public class MaxSumSubsequence { 4 private ArrayList<Integer> subseq; 5 public int getMaxSumSubseqNonAdj(int[] arr) { 6 if(arr == null || arr.length == 0) { 7 return 0; 8 } 9 int[] T = new int[arr.length + 1]; 10 T[0] = 0; T[1] = arr[0]; 11 for(int i = 2; i < T.length; i++) { 12 T[i] = Math.max(T[i - 1], T[i - 2] + arr[i - 1]); 13 } 14 subseq = new ArrayList<Integer>(); 15 int currSum = T[arr.length]; 16 int idx = arr.length; 17 while(currSum != 0) { 18 if(idx >= 2) { 19 if(T[idx - 1] <= T[idx - 2] + arr[idx - 1]) { 20 subseq.add(arr[idx - 1]); 21 currSum -= arr[idx - 1]; 22 idx -= 2; 23 } 24 else { 25 idx--; 26 } 27 } 28 else { 29 subseq.add(arr[idx - 1]); 30 currSum -= arr[idx - 1]; 31 idx--; 32 } 33 } 34 return T[arr.length]; 35 } 36 }
[Coding Made Simple] Maximum Sum Subsequence Non-adjacent
标签：null pre mat log other new dynamic lis imu
原文地址：http://www.cnblogs.com/lz87/p/7288855.html
