1286. Iterator for Combination

QUESTION:

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 is guaranteed that all calls of the function next are valid.

EXPLANATION:

提示中提到了用bit mask, 那么就用bitmask做一下, 其实简单的是用动态规划的回溯去做会比较简单一点.

SOLUTION:


class CombinationIterator {

        var result:[String] = []
        var pointer:Int = -1

        init(_ characters: String, _ combinationLength: Int) {
            var string:String = ""
            for i in 1...characters.count {
                string = string + "1"
            }
            var count:Int = Int(string, radix: 2)!
            var chars:[Character] = Array(characters)
            for i in 1...count {
                if i.nonzeroBitCount == combinationLength {
                    var mask:String = String(String(i, radix: 2).reversed())
                    var maskArr:[Character] = Array(mask)
                    var tmpResult:String = ""
                    for j in 0..<mask.count {
                        if (maskArr[j] == "1") {
                            tmpResult = tmpResult +  String(chars[j])
                        }
                    }
                    result.append(tmpResult)
                }
            }
            result = result.sorted()
        }
        
        func next() -> String {
            pointer += 1
            return result[pointer]
        }
        
        func hasNext() -> Bool {
            return pointer < result.count - 1
        }
}
>
--> return <--