标签：oge new inner initial control wro string together mat
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:
有n只队伍(n的范围[2, 212], 正偶数)，编号为1 ~ n，队伍按照最强和最弱的分在一组的原则分组比赛，给出一直到最后一轮比赛的分组方法。
解法1：迭代iterative，
解法2：递归recursive，
Java:
public String findContestMatch(int n) {
List<String> matches = new ArrayList<>();
for(int i = 1; i <= n; i++) matches.add(String.valueOf(i));
while(matches.size() != 1){
List<String> newRound = new ArrayList<>();
for(int i = 0; i < matches.size()/2; i++)
newRound.add("(" + matches.get(i) + "," + matches.get(matches.size() - i - 1) + ")");
matches = newRound;
}
return matches.get(0);
}
Java:
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];
}
Java: LinkedList
public String findContestMatch(int n) {
LinkedList<String> res = new LinkedList<>();
for (int i = 1; i <= n; i++) res.add(i + "");
while (res.size() > 1) {
LinkedList<String> tmp = new LinkedList<>();
while (!res.isEmpty()) {
tmp.add("(" + res.remove(0) + "," + res.remove(res.size() - 1) + ")");
}
res = tmp;
}
return res.get(0);
}
Java:
public string FindContestMatch(int n) {
string[] arr = new string[n];
for (int i = 0; i < n; i++) arr[i] = (i + 1).ToString();
int left = 0;
int right = n - 1;
while (left < right)
{
while (left < right)
{
arr[left] = "(" + arr[left] + "," + arr[right] + ")";
left++;
right--;
}
left = 0;
}
return arr[0];
}
Python:
# Time: O(n)
# Space: O(n)
class Solution(object):
def findContestMatch(self, n):
"""
:type n: int
:rtype: str
"""
matches = map(str, range(1, n+1))
while len(matches)/2:
matches = ["({},{})".format(matches[i], matches[-i-1]) for i in xrange(len(matches)/2)]
return matches[0]
Python:
class Solution(object):
def solve(self, groups):
size = len(groups)
if size == 1: return groups[0]
ngroups = []
for x in range(size / 2):
ngroups.append(‘(‘ + groups[x] + ‘,‘ + groups[size - x - 1] + ‘)‘)
return self.solve(ngroups)
def findContestMatch(self, n):
"""
:type n: int
:rtype: str
"""
return self.solve(map(str, range(1, n + 1)))
Python: wo
class Solution():
def contestMatches(self, n):
s = []
for i in xrange(n):
s.append(str(i + 1))
while n > 1:
cur = []
for m in xrange(len(s) / 2):
cur.append((s[m], s[len(s) - 1 - m]))
s = cur
n /= 2
return s[0]
Python: wo
class Solution():
def contestMatches(self, n):
s = []
for i in xrange(n):
s.append(str(i + 1))
return self.helper(s)
def helper(self, s):
if len(s) == 1:
return s[0]
curr = []
i, j = 0, len(s) - 1
while i < j:
curr.append((s[i], s[j]))
i += 1
j -= 1
return self.helper(curr)
C++:
// Time: O(n)
// Space: O(n)
class Solution {
public:
string findContestMatch(int n) {
vector<string> matches(n);
for (int i = 0; i < n; ++i) {
matches[i] = to_string(i + 1);
}
while (matches.size() / 2) {
vector<string> next_matches;
for (int i = 0; i < matches.size() / 2; ++i) {
next_matches.emplace_back("(" + matches[i] + "," + matches[matches.size() - 1 - i] + ")");
}
swap(matches, next_matches);
}
return matches[0];
}
};
C++:
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];
}
};
C++:
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);
}
};
[LeetCode] 544. Output Contest Matches 输出比赛匹配对
标签：oge new inner initial control wro string together mat
原文地址：https://www.cnblogs.com/lightwindy/p/9617310.html
