描述
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.
Note:
1 <= n <= 200
解析
根据题意，就是在对 n 做运算，那么就是将当前的 n 除 2 的结果加入 res 中，然后判断 如果 n 是偶数，直接将 n//2 的结果赋值给 n 进行后面的循环，如果 n 是奇数，则将 n//2+1 的结果赋值给 n 继续进行后面的循环即可。
解答
class Solution(object):
def numberOfMatches(self, n):
"""
:type n: int
:rtype: int
"""
res = 0
while n > 1:
res += n // 2
if n % 2 == 0:
n = n // 2
else:
n = n // 2 + 1
return res
运行结果
Runtime: 8 ms, faster than 99.19% of Python online submissions for Count of Matches in Tournament.
Memory Usage: 13.6 MB, less than 12.09% of Python online submissions for Count of Matches in Tournament.
原题链接：https://leetcode.com/problems/count-of-matches-in-tournament/
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