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Problem

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

Code

考虑两种情况

情况 1:

  • 1,2,3,4,5,6,7,8,9,10,11,12,13,14 - head, step, remain,
  • 1,2,3,4,5,6,7,8,9,10,11,12,13,14 - 1,1,14
  • ,2, ,4, ,6, ,8, ,10, ,12, ,14 - 2,2,7 ->
  • , , ,4, , , ,8, , , ,12, , - 4,4,3 <-
  • , , , , , , ,8, , , , , , - 8,8,1 ->

情况 2:

  • 1,2,3,4,5,6,7,8,9,10,11,12,13 - head, step, remain,
  • 1,2,3,4,5,6,7,8,9,10,11,12,13 - 1,1,13
  • ,2, ,4, ,6, ,8, ,10, ,12, - 2,2,6 ->
  • ,2, , , ,6, , , ,10, , , - 2,4,3 <-
  • , , , , ,6, , , , , , , - 6,8,1 ->

结论:

  • 从左删除的时候: 每次都要让 head + step
  • 从右删除的时候:
    • 如果剩下奇数个数字, head + step, 比如 2, 4, 6
    • 如果剩下偶数个数字, head 不变, 比如 2, 4, 6, 8

每次删除, 数字之间的距离都会增加一倍, 因此 step 每次都乘以 2

class Solution {
    public int lastRemaining(int n) {
        int head = 1;
        int remain = n;
        int step = 1;
        boolean isLeft = true;

        while(remain != 1){
            // 从left || 从right, 但有奇数个元素
            if(isLeft || (!isLeft && remain % 2 == 1)){
                head += step;
            }
            remain /= 2;
            step *= 2;
            isLeft = !isLeft;
        }

        return head;
    }
}