544. Output Contest Matches

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))).

Solution：

思路：
1 2 3 4 5 6 7 8
因为是最强和最弱来对决，其次是次强与次弱对决，以此类推可得到：
1-8 2-7 3-6 4-5
那么接下来呢，还是最强与最弱，次强与次弱这种关系：
(1-8 4-5) (2-7 3-6)
最后胜者争夺冠军
((1-8 4-5) (2-7 3-6))

由于n限定了是2的次方数，那么就是可以一直对半分的，比如开始有n队，第一拆分为n/2对匹配，然后再对半拆，就是n/2/2，直到拆到n为1停止，而且每次都是首与末配对，次首与次末配对.

Time Complexity: O(N) Space Complexity: O(N)

Solution Code：

public class Solution {
    public String findContestMatch(int n) {
        String[] m = new String[n];
        for (int i = 0; i < n; i++) {
            m[i] = String.valueOf(i + 1);
        }

        while (n > 1) {
            for (int i = 0; i < n / 2; i++) {
                m[i] = "(" + m[i] + "," + m[n - 1 - i] + ")";
            }
            n /= 2;
        }
        
        return m[0];
    }
}