LeetCode 1286. Iterator for Combination

原题链接在这里：https://leetcode.com/problems/iterator-for-combination/

题目：

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.

题解：

It comes to how to find the next combination. Say characters = "abcd", current combination = "acd".

1. Remove the common suffix with characters, which is "cd".
2. Now current String = "a". Remove the right most char, which is 'a'. current String = "". Find the index of a in characters, which is 0.
3. Increment that index by 1, 0++. It points to 'b'. Append this char and the following until current combination length = combinationLength.

Could use StringBuilder to maintain the current combination.

Time Complexity: CombinationIterator, O(n). next, O(n). hasNext, O(1). n = characters.length().

Space: O(n).

AC Java:

class CombinationIterator {
    StringBuilder sb;
    String s;
    int len;
    HashMap<Character, Integer> hm;

    public CombinationIterator(String characters, int combinationLength) {
        this.sb = new StringBuilder();
        this.s = characters;
        this.len = combinationLength;
        this.hm = new HashMap<>();
        
        for(int i = 0; i<s.length(); i++){
            char c = s.charAt(i);
            if(sb.length() < len){
                sb.append(c);
            }
            
            hm.put(c, i);
        }
    }
    
    public String next() {
        String res = sb.toString();
        int ind = s.length() - 1;
        while(ind >= 0 && sb.length() != 0 && sb.charAt(sb.length() -1) == s.charAt(ind)){
            sb.deleteCharAt(sb.length() - 1);
            ind--;
        }
        
        if(sb.length() == 0){
            return res;
        }
        
        char c = sb.charAt(sb.length() - 1);
        ind = hm.get(c) + 1;
        sb.deleteCharAt(sb.length() - 1);
        
        while(sb.length() < len && ind < s.length()){
            sb.append(s.charAt(ind));
            ind++;
        }
        
        return res;
    }
    
    public boolean hasNext() {
        return sb.length() != 0;
    }
}

/**
 * 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();
 */

类似Next Permutation.