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80 lines
2.5 KiB
Markdown
80 lines
2.5 KiB
Markdown
# [1688. Count of Matches in Tournament](https://leetcode.com/problems/count-of-matches-in-tournament/)
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## 题目
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You are given an integer `n`, the number of teams in a tournament that has strange rules:
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- 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.
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- 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.
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Return *the number of matches played in the tournament until a winner is decided.*
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**Example 1:**
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```
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Input: n = 7
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Output: 6
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Explanation: Details of the tournament:
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- 1st Round: Teams = 7, Matches = 3, and 4 teams advance.
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- 2nd Round: Teams = 4, Matches = 2, and 2 teams advance.
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- 3rd Round: Teams = 2, Matches = 1, and 1 team is declared the winner.
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Total number of matches = 3 + 2 + 1 = 6.
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```
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**Example 2:**
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```
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Input: n = 14
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Output: 13
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Explanation: Details of the tournament:
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- 1st Round: Teams = 14, Matches = 7, and 7 teams advance.
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- 2nd Round: Teams = 7, Matches = 3, and 4 teams advance.
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- 3rd Round: Teams = 4, Matches = 2, and 2 teams advance.
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- 4th Round: Teams = 2, Matches = 1, and 1 team is declared the winner.
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Total number of matches = 7 + 3 + 2 + 1 = 13.
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```
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**Constraints:**
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- `1 <= n <= 200`
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## 题目大意
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给你一个整数 n ，表示比赛中的队伍数。比赛遵循一种独特的赛制：
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- 如果当前队伍数是 偶数 ，那么每支队伍都会与另一支队伍配对。总共进行 n / 2 场比赛，且产生 n / 2 支队伍进入下一轮。
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- 如果当前队伍数为 奇数 ，那么将会随机轮空并晋级一支队伍，其余的队伍配对。总共进行 (n - 1) / 2 场比赛，且产生 (n - 1) / 2 + 1 支队伍进入下一轮。
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返回在比赛中进行的配对次数，直到决出获胜队伍为止。
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## 解题思路
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- 简单题，按照题目的规则模拟。
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- 这一题还有更加简洁的代码，见解法一。n 个队伍，一个冠军，需要淘汰 n-1 个队伍。每一场比赛淘汰一个队伍，因此进行了 n-1 场比赛。所以共有 n-1 个配对。
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## 代码
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```go
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package leetcode
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// 解法一
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func numberOfMatches(n int) int {
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return n - 1
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}
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// 解法二 模拟
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func numberOfMatches1(n int) int {
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sum := 0
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for n != 1 {
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if n&1 == 0 {
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sum += n / 2
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n = n / 2
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} else {
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sum += (n - 1) / 2
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n = (n-1)/2 + 1
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}
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}
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return sum
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}
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``` |