LeetCode #1286 Iterator for Comb
1286 Iterator for Combination 字母组合迭代器
Description:
Design the CombinationIterator class:
CombinationIterator(string characters, int combinationLength) Initializes the object with a string characters of sorted distinct lowercase English letters and a number combinationLength as arguments.
next() Returns the next combination of length combinationLength in lexicographical order.
hasNext() Returns true if and only if there exists a next combination.
Example:
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 10^4 calls will be made to next and hasNext.
It is guaranteed that all calls of the function next are valid.
题目描述:
请你设计一个迭代器类 CombinationIterator ,包括以下内容:
CombinationIterator(string characters, int combinationLength) 一个构造函数,输入参数包括:用一个 有序且字符唯一 的字符串 characters(该字符串只包含小写英文字母)和一个数字 combinationLength 。
函数 next() ,按 字典序 返回长度为 combinationLength 的下一个字母组合。
函数 hasNext() ,只有存在长度为 combinationLength 的下一个字母组合时,才返回 true
示例:
示例 1:
输入:
["CombinationIterator", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
[["abc", 2], [], [], [], [], [], []]
输出:
[null, "ab", true, "ac", true, "bc", false]
解释:
CombinationIterator iterator = new CombinationIterator("abc", 2); // 创建迭代器 iterator
iterator.next(); // 返回 "ab"
iterator.hasNext(); // 返回 true
iterator.next(); // 返回 "ac"
iterator.hasNext(); // 返回 true
iterator.next(); // 返回 "bc"
iterator.hasNext(); // 返回 false
提示:
1 <= combinationLength <= characters.length <= 15
characters 中每个字符都 不同
每组测试数据最多对 next 和 hasNext 调用 10^4次
题目保证每次调用函数 next 时都存在下一个字母组合。
思路:
回溯
DFS 预处理所有的字符串到一个列表中
直接用列表返回结果
也可以使用生成法
第一个组合为初始字符串的前 k 个字符
从第 k 个字符开始, 如果有没使用过的比当前位置更大的字符, 直接替换
否则往前寻找第一个可以替换的字符, 并将该位置到 k 位置上的字符都换成最小的
时间复杂度为 O(k), 空间复杂度为 O(1)
代码:
C++:
class CombinationIterator
{
private:
vector<string> list;
string cur = "";
int index = 0;
void dfs(string& characters, int combinationLength, int index)
{
if (cur.size() == combinationLength) list.emplace_back(cur);
for (int i = index; i < characters.length(); ++i)
{
cur += characters[i];
dfs(characters, combinationLength, i + 1);
cur.erase(cur.size() - 1);
}
}
public:
CombinationIterator(string characters, int combinationLength)
{
dfs(characters, combinationLength, 0);
}
string next()
{
return list[index++];
}
bool hasNext()
{
return index < list.size();
}
};
/**
* Your CombinationIterator object will be instantiated and called as such:
* CombinationIterator obj = new CombinationIterator(characters, combinationLength);
* String param_1 = obj.next();
* boolean param_2 = obj.hasNext();
*/
/**
* Your CombinationIterator object will be instantiated and called as such:
* CombinationIterator* obj = new CombinationIterator(characters, combinationLength);
* string param_1 = obj->next();
* bool param_2 = obj->hasNext();
*/
Java:
class CombinationIterator {
private List<String> list = new LinkedList<>();
private StringBuilder sb = new StringBuilder();
private int index = 0;
public CombinationIterator(String characters, int combinationLength) {
dfs(characters, combinationLength, 0);
}
private void dfs(String characters, int combinationLength, int index){
if (sb.length() == combinationLength) list.add(sb.toString());
for (int i = index; i < characters.length(); i++) {
sb.append(characters.charAt(i));
dfs(characters, combinationLength, i + 1);
sb.deleteCharAt(sb.length() - 1);
}
}
public String next() {
return list.get(index++);
}
public boolean hasNext() {
return index < list.size();
}
}
/**
* Your CombinationIterator object will be instantiated and called as such:
* CombinationIterator obj = new CombinationIterator(characters, combinationLength);
* String param_1 = obj.next();
* boolean param_2 = obj.hasNext();
*/
Python:
class CombinationIterator:
def __init__(self, characters: str, combinationLength: int):
self.data = list(combinations(characters, combinationLength))
self.pos = 0
self.n = len(self.data)
def next(self) -> str:
result = ''.join(self.data[self.pos])
self.pos += 1
return result
def hasNext(self) -> bool:
return self.pos < self.n
# Your CombinationIterator object will be instantiated and called as such:
# obj = CombinationIterator(characters, combinationLength)
# param_1 = obj.next()
# param_2 = obj.hasNext()