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[LeetCode] 1688. Count of Matches in Tournament

发布时间 2023-12-06 06:00:33作者: CNoodle

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

比赛中的配对次数。

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：
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。
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。
返回在比赛中进行的配对次数，直到决出获胜队伍为止。

思路

当队伍个数 n 不等于 1 的时候，就需要一直配对来比较到底谁是最后的胜者。题目规定了当队伍个数为偶数的时候，就两两配对看谁是胜者；当队伍个数为偶数的时候，可以随机挑出一个胜者，让剩下的人再两两配对看谁是胜者，所以无论队伍个数是奇数还是偶数，配对次数永远都是当前队伍个数 / 2。

复杂度

时间O(n/2)
空间O(1)

代码

Java实现

class Solution {
    public int numberOfMatches(int n) {
        int count = 0;
        while (n != 1) {
            count += n / 2;
            if (n % 2 == 1) {
                n = n / 2 + 1;
            } else {
                n /= 2;
            }
        }
        return count;
    }
}