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【leetcode】1688. Count of Matches in Tournament

2022-02-04 16:33:00  阅读:193  来源: 互联网

标签:Count advance Matches Tournament number Teams 1688 teams Round


题目如下:

You are given an integer n, the number of teams in a tournament that has strange rules:

  • If the current number of teams is even, each team gets paired with another team. A total of n / 2 matches are played, and n / 2 teams advance to the next round.
  • If the current number of teams is odd, one team randomly advances in the tournament, and the rest gets paired. A total of (n - 1) / 2 matches are played, and (n - 1) / 2 + 1 teams advance to the next round.

Return the number of matches played in the tournament until a winner is decided.

 

Example 1:

Input: n = 7
Output: 6
Explanation: Details of the tournament: 
- 1st Round: Teams = 7, Matches = 3, and 4 teams advance.
- 2nd Round: Teams = 4, Matches = 2, and 2 teams advance.
- 3rd Round: Teams = 2, Matches = 1, and 1 team is declared the winner.
Total number of matches = 3 + 2 + 1 = 6.

Example 2:

Input: n = 14
Output: 13
Explanation: Details of the tournament:
- 1st Round: Teams = 14, Matches = 7, and 7 teams advance.
- 2nd Round: Teams = 7, Matches = 3, and 4 teams advance.
- 3rd Round: Teams = 4, Matches = 2, and 2 teams advance.
- 4th Round: Teams = 2, Matches = 1, and 1 team is declared the winner.
Total number of matches = 7 + 3 + 2 + 1 = 13. 

Constraints:

  • 1 <= n <= 200

解题思路:很简单的题目,直接算吧。

代码如下:

class Solution(object):
    def numberOfMatches(self, n):
        """
        :type n: int
        :rtype: int
        """
        res = 0
        while n > 1:
            if n == 2:
                res += 1
                break
            elif n % 2 == 0:
                res += n / 2
                n /= 2
            else:
                res += n / 2
                n = n/2 + 1
        return res

 

标签:Count,advance,Matches,Tournament,number,Teams,1688,teams,Round
来源: https://www.cnblogs.com/seyjs/p/15862979.html

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