[LeetCode] Output Contest Matches 輸出比賽匹配對

 

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:

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

 

這道題講的是NBA的季后賽對戰順序，對於一個看了十幾年NBA的老粉來說，再熟悉不過了。這種對戰順序是為了避免強強之間過早對決，從而失去比賽的公平性，跟歐冠歐聯那種八強就開始隨機抽簽匹配有本質上的區別。NBA的這種比賽機制基本弱隊很難翻身，假如你是拿到最后一張季后賽門票進的，那么一上來就干聯盟第一，肯定凶多吉少，很有可能就被橫掃了。但是偶爾也會出現黑八的情況，但都是極其少見的，畢竟像勇士這么叼的球隊畢竟不多。好了，不閑扯了，來做題吧。我們就拿NBA這種八個球隊的情況來分析吧，八只球隊的排名是按常規賽勝率來排的：

1 2 3 4 5 6 7 8

因為是最強和最弱來對決，其次是次強與次弱對決，以此類推可得到：

1-8  2-7  3-6  4-5

那么接下來呢，還是最強與最弱，次強與次弱這種關系：

(1-8  4-5)  (2-7  3-6)

最后勝者爭奪冠軍

((1-8  4-5)  (2-7  3-6))

這樣一分析是不是就清楚了呢，由於n限定了是2的次方數，那么就是可以一直對半分的，比如開始有n隊，第一拆分為n/2對匹配，然后再對半拆，就是n/2/2，直到拆到n為1停止，而且每次都是首與末配對，次首與次末配對，這樣搞清楚了規律，代碼應該就不難寫了吧，參見代碼如下：

 

解法一：

class Solution {
public:
    string findContestMatch(int n) {
        vector<string> v;
        for (int i = 1; i <= n; ++i) v.push_back(to_string(i));
        while (n > 1) {
            for (int i = 0; i < n / 2; ++i) {
                v[i] = "(" + v[i] + "," + v[n - i - 1] + ")";
            }
            n /= 2;
        }
        return v[0];
    }
};

 

下面這種方法是遞歸的寫法，解題思路跟上面沒有區別，參見代碼如下：

 

解法二：

class Solution {
public:
    string findContestMatch(int n) {
        vector<string> v;
        for (int i = 1; i <= n; ++i) v.push_back(to_string(i));
        helper(n, v);
        return v[0];
    }
    void helper(int n, vector<string>& v) {
        if (n == 1) return;
        for (int i = 0; i < n; ++i) {
            v[i] = "(" + v[i] + "," + v[n - i - 1] + ")";
        }
        helper(n / 2, v);
    }
};

 

參考資料：

https://discuss.leetcode.com/topic/83454/java-10-lines/2

https://discuss.leetcode.com/topic/83457/c-java-clean-code

https://discuss.leetcode.com/topic/83460/java-recursive-solution

 

LeetCode All in One 題目講解匯總(持續更新中...)