There is a list of sorted integers from 1 to n. Starting from left to right, remove the first number and every other number afterward until you reach the end of the list.
Repeat the previous step again, but this time from right to left, remove the right most number and every other number from the remaining numbers.
We keep repeating the steps again, alternating left to right and right to left, until a single number remains.
Find the last number that remains starting with a list of length n.
Example:
Input: n = 9, 1 2 3 4 5 6 7 8 9 2 4 6 8 2 6 6 Output: 6
这道题dp的方法如下，dp能比较好的显示思路
public int LastRemaining(int n) { if(n<=2) return n; if(n==3) return 2; var left = new int[n+1]; var right = new int[n+1]; left[1] = 1; right[1] = 1; left[2] = 2; right[2] = 1; for(int i= 3;i<=n;i++) { if(i%2 == 0) { left[i] = 2*right[i/2]; right[i] = 2*left[i/2]-1; } else { left[i] = 2*right[i/2]; right[i] = 2*left[i/2]; } } return left[n]; }
然后DP需要的space太大了，所以这题适合用backtracking做
public int LastRemaining(int n) { return BackTracking(n,true); } private int BackTracking(int n, bool left) { if(n<=2) return left?n:1; if(n%2 == 0) return 2*BackTracking(n/2, !left)-(left?0:1); else return 2*BackTracking(n/2,!left); }
原文：http://www.cnblogs.com/renyualbert/p/5888867.html