[leetcode] 390. Elimination Game

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 92 4 6 82 66Output:6

这道题是数字消除游戏，题目难度为Medium。

我们用head表示当前数字序列的第一个数字，gap表示当前数字序列两数字之间的间隔，经过多次迭代最终剩下数字个数为1时head即是要找的数字。如果从左向右删除，则第一个数字必定会删除，因此需要更新head，向前移动gap个数字；如果从右向左删除，数字个数为偶数时第一个数字不会被删除，head不变，数字个数为奇数时第一个数字会被删除，head同样需要向前移动gap个数字。gap从1开始每删除一轮数字翻倍。这样迭代直至数字个数变为1即可得出最终剩下的数字。具体代码：

class Solution {public:    int lastRemaining(int n) {        int head = 1, gap = 1;        bool left2right = true;        while(n != 1) {            if(left2right || n%2) head += gap;            gap *= 2;            n /= 2;            left2right = !left2right;        }        return head;    }};