标签：const iterator etc ash set cal mos ade style
问题：
设计一个组合指针类，通过给定组合元素characters，要求构成组合的大小combinationLength，实现以下方法
构造方法：CombinationIterator(string characters, int combinationLength)
返回下一个组合：next()
是否存在下一个组合：hasNext()
Example 1:
Input
["CombinationIterator", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
[["abc", 2], [], [], [], [], [], []]
Output
[null, "ab", true, "ac", true, "bc", false]
Explanation
CombinationIterator itr = new CombinationIterator("abc", 2);
itr.next(); // return "ab"
itr.hasNext(); // return True
itr.next(); // return "ac"
itr.hasNext(); // return True
itr.next(); // return "bc"
itr.hasNext(); // return False
Constraints:
1 <= combinationLength <= characters.length <= 15
All the characters of characters are unique.
At most 104 calls will be made to next and hasNext.
It‘s guaranteed that all calls of the function next are valid.
解法：
保存构造体：vector<string>
解法一：Backtracking（回溯算法）
代码参考：
1 class CombinationIterator { 2 public: 3 vector<string> comb; 4 int itr; 5 void dfs(string characters, int combinationLength, int pos, string path) { 6 if(combinationLength == path.length()) { 7 comb.push_back(path); 8 return; 9 } 10 for(int i=pos; i<characters.size(); i++) { 11 dfs(characters, combinationLength, i+1, path+characters[i]); 12 } 13 } 14 CombinationIterator(string characters, int combinationLength) { 15 itr=0; 16 dfs(characters, combinationLength, 0, ""); 17 } 18 19 string next() { 20 return itr!=comb.size()?comb[itr++]:""; 21 } 22 23 bool hasNext() { 24 return itr!=comb.size(); 25 } 26 }; 27 28 /** 29 * Your CombinationIterator object will be instantiated and called as such: 30 * CombinationIterator* obj = new CombinationIterator(characters, combinationLength); 31 * string param_1 = obj->next(); 32 * bool param_2 = obj->hasNext(); 33 */
解法二：
保存构造体：set<string>（由题意，有顺序要求）
bit组合法：
每个元素代表一个bit位，选择表示1，不选表示0
那么，题意中要求选择长度combinationLength的元素构成组合，即：选择所有元素构成bit中含有2个1的组合。
再将这两个1对应的元素拿出来，拼成解，加入解list中。
Suppose Input string - "abc" and combinationLength =2
now mask = 8 which is nothing but 2 power length of of "abc"
now Itarate from 1 to 8
Num bit_rep no_of_set_bit a b c hold_set
1 0 0 1 1 0 0 1 nothing <= because combinationLength != no_of_set_bit
2 0 1 0 1 0 1 0 same case like upper one
3 0 1 1 2 0 1 1 "bc" <= combinationLength ==no_of
4 1 0 0 1 1 0 0 nothing
5 1 0 1 2 1 0 1 "ac"
6 1 1 0 2 1 1 0 "ab"
7 1 1 1 3 1 1 1 nothing <= because combinationLength != no_of_set_bit
return that set in lexi order ["ab","ac","bc"]
Now We Done Our Main Work
代码参考：
1 class CombinationIterator { 2 public: 3 //vector<string> comb; 4 set<string> comb; 5 //int itr; 6 set <string> :: iterator itr; 7 /*void dfs(string characters, int combinationLength, int pos, string path) { 8 if(combinationLength == path.length()) { 9 comb.push_back(path); 10 return; 11 } 12 for(int i=pos; i<characters.size(); i++) { 13 dfs(characters, combinationLength, i+1, path+characters[i]); 14 } 15 }*/ 16 void creatcomb(string s, int len) { 17 int maxn = 1<<s.length(); 18 for(int cmb=1; cmb<maxn; cmb++) { 19 //count the num of 1: 20 if(__builtin_popcount(cmb)!=len) continue; 21 string newcomb; 22 //the bit 1 in cmb means the alphbet in s would be selected. 23 for(int i=cmb, idx=0; i>0; i>>=1, idx++) { 24 if(i&1) newcomb+=s[idx]; 25 } 26 comb.insert(newcomb); 27 } 28 return; 29 } 30 CombinationIterator(string characters, int combinationLength) { 31 //itr=0; 32 creatcomb(characters, combinationLength); 33 itr=comb.begin(); 34 //dfs(characters, combinationLength, 0, ""); 35 } 36 37 string next() { 38 return itr!=comb.end()?*itr++:""; 39 } 40 41 bool hasNext() { 42 return itr!=comb.end(); 43 } 44 }; 45 46 /** 47 * Your CombinationIterator object will be instantiated and called as such: 48 * CombinationIterator* obj = new CombinationIterator(characters, combinationLength); 49 * string param_1 = obj->next(); 50 * bool param_2 = obj->hasNext(); 51 */
1286. Iterator for Combination
标签：const iterator etc ash set cal mos ade style
原文地址：https://www.cnblogs.com/habibah-chang/p/14349890.html
