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?yàn)槭亲顝?qiáng)和最弱來對(duì)決，其次是次強(qiáng)與次弱對(duì)決躺屁，以此類推可得到：
1-8 2-7 3-6 4-5
那么接下來呢答毫，還是最強(qiáng)與最弱畦粮，次強(qiáng)與次弱這種關(guān)系：
(1-8 4-5) (2-7 3-6)
最后勝者爭奪冠軍
((1-8 4-5) (2-7 3-6))
由于n限定了是2的次方數(shù)绞吁，那么就是可以一直對(duì)半分的哗脖，比如開始有n隊(duì)糊肠，第一拆分為n/2對(duì)匹配，然后再對(duì)半拆悯辙，就是n/2/2琳省，直到拆到n為1停止，而且每次都是首與末配對(duì)笑撞，次首與次末配對(duì).
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];
}
}
