leetcode 1286. Iterator for Combination
Design an Iterator class, which has:
- A constructor that takes a string
characters
of sorted distinctlowercase English letters and a numbercombinationLength
as arguments. - A function next() that returns the next combination of length
combinationLength
in lexicographical order. - A function hasNext() that returns
True
if and only if there exists a next combination.
python3:利用itertools.combinations模块
1 class CombinationIterator: 2 3 def __init__(self, A: str, k: int): 4 self.it = itertools.combinations(A, k) 5 self.last = A[-k:] 6 self.res = "" 7 8 9 def next(self) -> str: 10 self.res = ''.join(next(self.it)) 11 return self.res 12 13 def hasNext(self) -> bool: 14 return self.res != self.last
C++:先产生所有符合字典序的子字符串。(回溯法)
1 class CombinationIterator { 2 private: 3 queue<string> iterator; 4 void combinationIterator(const string &characters, const int &combinationLength, int startpos, string &oneInstance, queue<string> &resultQueue) { 5 if (oneInstance.length() == combinationLength) { 6 resultQueue.push(oneInstance); 7 return; 8 } 9 for (int i = startpos; i < characters.length(); ++i) { 10 oneInstance.push_back(characters[i]); 11 combinationIterator(characters, combinationLength, i + 1, oneInstance, resultQueue); 12 oneInstance.pop_back(); 13 } 14 } 15 public: 16 CombinationIterator(string characters, int combinationLength) { 17 string oneInstance = ""; 18 combinationIterator(characters, combinationLength, 0, oneInstance, iterator); 19 } 20 21 string next() { 22 string ans = ""; 23 if (hasNext()) { 24 ans = iterator.front(); 25 iterator.pop(); 26 } 27 return ans; 28 } 29 30 bool hasNext() { 31 //队列非空,说明还有 32 return !iterator.empty(); 33 } 34 };
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