【leetcode】390. Elimination Game

The topics are as follows:

Problem- solving ideas: For this kind of number type of questions, the numbers generally have inherent rules. No matter how many times you operate, the array of this question is always an arithmetic sequence. From the sequence [1 2 3 4 5 6 7 8 9] -> [2 4 6 8] -> [2 6] -> [6], we can get the tolerances to be 1, 2, and 4, respectively. If we expand n a little and print out the remaining array sequence of each step, we can easily find that the tolerance is the power of pow(2,n). Next, we only need to record the low, high and the length of the remaining sequence after each operation, until the last sequence has only one element.

code show as below:

class Solution(object):
    def lastRemaining(self, n):
        """
        :type n: int
        :rtype: int
        """
        if n == 1:
            return 1
        times = 1
        low = high = None
        length = n
        multiple = None
        while True:
            if times == 1:
                length = length / 2
                low = 2
                if n % 2 == 0:
                    high = n
                else:
                    high = n -1
                multiple = pow(2, times)
            elif times % 2 == 0:
                length = length / 2
                high -= multiple
                multiple = pow(2, times)
                low = high - multiple*(length-1)
            else:
                length = length / 2
                low += multiple
                multiple = pow(2, times)
                high = low + multiple * (length - 1)
            times += 1
            if low >= high:
                return high