【leetcode】544. Output Contest Matches

原题

During the NBA playoffs, we always arrange the rather strong team to play with the rather weak team,
like make the rank 1 team play with the rank nth team, which is a good strategy
to make the contest more interesting. Now, you're given n teams, you need to output their final contest matches
in the form of a string.
The n teams are given in the form of positive integers from 1 to n, which represents their initial rank.
(Rank 1 is the strongest team and Rank n is the weakest team.) We'll use parentheses('(', ')') and commas(',')
to represent the contest team pairing - parentheses('(' , ')') for pairing and commas(',') for partition.
During the pairing process in each round, you always need to follow the strategy of making
the rather strong one pair with the rather weak one.
Example 1:
Input: 2
Output: (1,2)
Explanation:
Initially, we have the team 1 and the team 2, placed like: 1,2.
Then we pair the team (1,2) together with '(', ')' and ',', which is the final answer.

Example 2:
Input: 4
Output: ((1,4),(2,3))
Explanation:
In the first round, we pair the team 1 and 4, the team 2 and 3 together, as we need to make the strong team and weak team together.
And we got (1,4),(2,3).
In the second round, the winners of (1,4) and (2,3) need to play again to generate the final winner, so you need to add the paratheses outside them.
And we got the final answer ((1,4),(2,3)).

Example 3:
Input: 8
Output: (((1,8),(4,5)),((2,7),(3,6)))
Explanation:
First round: (1,8),(2,7),(3,6),(4,5)
Second round: ((1,8),(4,5)),((2,7),(3,6))
Third round: (((1,8),(4,5)),((2,7),(3,6)))
Since the third round will generate the final winner, you need to output the answer (((1,8),(4,5)),((2,7),(3,6))).
Note:

The n is in range [2, 212].
We ensure that the input n can be converted into the form 2k, where k is a positive integer.

解析

输出比赛配对
最强和最弱匹配，次强和次若匹配，一轮后，继续按此规则配对，输出最终的配对规则

思路

递归

解法

    public String findContestMatch(int n) {
        List<String> teams = new ArrayList<>();
        for (int i = 1; i <= n; i++) {
            teams.add(String.valueOf(i));
        }
        return match(teams);
    }

    private String match(List<String> teams) {
        List<String> newTeams = new ArrayList<>();
        for (int i = 0; i < teams.size() / 2; i++) {
            newTeams.add(String.format("(%s,%s)", teams.get(i), teams.get(teams.size() - i - 1)));
        }
        if (newTeams.size() > 1) {
            return match(newTeams);
        }
        return newTeams.get(0);
    }