390. Elimination Game -------M

 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

A:

就是分开分析， 然后就得到最优解了。

class Solution {
public:
    int lastRemaining(int n) {
        return left2right(n);
    }
private:
    int left2right(int n){
        if(n ==1)
            return 1;
        return 2*(right2left(n/2));
    }
private:
    int right2left(int n){
        if(n==1)
            return 1;
        if(n%2==1){ // assume 1,2,3
            return 2*left2right(n/2);
        }else{ // assume 1,2,3,4,  we will have 1,3 left，(2,4)-1-->2*(1,2)-1
            return 2* left2right(n/2)-1;
        }
    }
};

上面的lastRemaining和left2right其实是重复的。因此可以去掉left2right

class Solution {
public:
    int lastRemaining(int n) {
        if(n ==1)
            return 1;
        return 2* right2left(n/2);
    }
private:
    int right2left(int n){
        if(n==1)
            return 1;
        if(n%2==1){ // assume 1,2,3
            return 2*lastRemaining(n/2);
        }else{ // assume 1,2,3,4,  we will have 1,3 left
            return 2* lastRemaining(n/2)-1;
        }
    }
};