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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 = 7Output: 6Explanation: 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 = 14Output: 13Explanation: 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

题意：一个整数 n 表示比赛中的队伍数。返回在比赛中进行的配对次数，直到决出获胜队伍为止。

解法1 遵循题意

class Solution {
   public:    int numberOfMatches(int n) {
           int ans = 0;        while (n > 1) {
    //n<=1时无需比赛            if (n & 1) {
    ans += (n - 1) / 2; n = (n - 1) / 2 + 1; } //奇数,加上(n-1)/2次配对次数,剩下(n-1)/2+1队伍            else {
    ans += n / 2; n /= 2; } //偶数,加上n/2次配对次数,剩下n/2支队伍        }        return ans;    }};

运行效率如下：

执行用时：0 ms, 在所有 C++ 提交中击败了100.00% 的用户内存消耗：6.2 MB, 在所有 C++ 提交中击败了100.00% 的用户

解法2 数学规律

每一轮中会进行多次配对，每次配对进行比赛，都会淘汰一支队伍。直到决出获胜队伍。因此最终只进行 n - 1 次配对：

class Solution {
   public:    int numberOfMatches(int n) {
           return n - 1;    }};

运行效率如下：

执行用时：0 ms, 在所有 C++ 提交中击败了100.00% 的用户内存消耗：6.2 MB, 在所有 C++ 提交中击败了100.00% 的用户

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