[LeetCode] 1286. Iterator for Combination

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 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.

字母组合迭代器。

请你设计一个迭代器类，包括以下内容：

一个构造函数，输入参数包括：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；否则，返回 False。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/iterator-for-combination
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这是一道回溯的题目，如果不熟悉可以先做77题。77题的组合问题是给你从 1 到 N 的数字请你列出所有长度为 K 的数字的组合，这道题无非是把数字换成了字母。注意题目条件说了input字符串中字母是有序且唯一的，所以我们就无需排序和处理重复结果的问题了。

时间O(n)

空间O(n)

Java实现

class CombinationIterator {
    Queue<String> queue;

    public CombinationIterator(String characters, int combinationLength) {
        queue = new LinkedList<>();
        helper(characters, 0, "", combinationLength, queue);
    }

    public void helper(String characters, int start, String cur, int k, Queue<String> queue) {
        // k = how may characters left for current string, k = combinationLength - soFar.length()
        if (k == 0) {
            queue.offer(cur);
            return;
        }
        for (int i = start; i < characters.length(); i++) {
            helper(characters, i + 1, cur + characters.charAt(i), k - 1, queue);
        }
    }
    
    public String next() {
        return queue.poll();
    }
    
    public boolean hasNext() {
        return !queue.isEmpty();
    }
}

/**
 * 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();
 */

相关题目

77. Combinations

1286. Iterator for Combination

LeetCode 题目总结