879. Output Contest Matches
Description
中文English
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.- The
nis in range[2, 2^12]. - We ensure that the input
ncan be converted into the form2^k, where k is a positive integer.
Have you met this question in a real interview?
Example
Given n =
2, return "(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.
Given n =
4, return "((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)).
Given n =
8, return "(((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))).
- DifficultyMedium
- Total Accepted390
- Total Submitted625
- Accepted Rate62%
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string findContestMatch(int n) { // write your code here //整体的结构： 8 -> 4 -> 2-> 1，每次砍一半 std::vector<string> match; for(int i = 1; i <= n; i++){ match.push_back(to_string(i)); } /* while(n > 1){ for(int i = 0; i < n / 2; i++){ match[i] = "(" + match[i] + "," + match[n - i - 1] + ")"; } n /= 2; } */ // 把这里改为recursion的方式 helper(n, match); return match[0]; } void helper(int len, vector<string> &match){ if(len == 1){ return; } for(int i = 0; i < len/2; i++){ match[i] = "(" + match[i] + "," + match[len - i - 1] + ")"; } helper(len / 2, match); } };
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