打印出字符串的所有组合 [英] Printing out all the combinations of a string

问题描述

iOS 11，Swift 4.0

iOS 11, Swift 4.0

尝试编写一个递归函数以显示字符串的所有可能组合.我知道了，但由于我只得到20对，我应该得到24对，所以不太正确.我看不到我在这里错过的一切.

Trying to write a recursive function to show all the possible combinations of a string. I got this, but its not quite right since I get only 20 pairs and I should get 24. I cannot see what I have missed here.

该编码在哪里出错?

var ans:Set<String>!

override func viewDidLoad() {
    super.viewDidLoad()
    // Do any additional setup after loading the view, typically from a nib.
    let str = "ABCD"
    ans = []
    recursiveString(s2s: str, char2s: 0)
    print("\(ans) \(ans.count)")
}

func recursiveSwap(s2x: String, c2x: Int, j2m: Int) {
    var anschr = Array(s2x)
        let tmpchr = anschr[c2x]
        anschr[c2x] = anschr[c2x+j2m]
        anschr[c2x+j2m] = tmpchr
        print("\(String(anschr))")
            ans.insert(String(anschr))
        if (c2x + j2m + 1) < s2x.count {
            recursiveSwap(s2x: String(s2x), c2x: c2x, j2m: j2m+1)
        } else {
            if (c2x + 1) < s2x.count - 1 {
                recursiveSwap(s2x: String(anschr), c2x: c2x + 1, j2m: 1)
            }
        }
}

func recursiveString(s2s: String, char2s: Int) {
    let blue = shiftString(s2s: s2s)
    if char2s < s2s.count {
        recursiveSwap(s2x: blue, c2x: 0, j2m: 1)
        recursiveString(s2s: blue, char2s: char2s + 1)
    }
}

func shiftString(s2s: String) -> String {
    let str2s = Array(s2s)
    let newS = str2s.suffix(str2s.count - 1)  + str2s.prefix(1)
    return String(newS)
}

给我...

CBDA DCBA 交流数据库 ADCB ABDC A B C D DCAB ADCB 数模转换器 南部非洲发展共同体 计算机辅助设计 BCDA 公元前 南部非洲发展共同体 中央商务区 工商管理学院 CDBA CDAB BACD 工商管理学院 DBCA DCBA 数模转换器 DABC

CBDA DCBA ACDB ADCB ABDC ABCD DCAB ADCB BDAC BADC BCAD BCDA ADBC BADC CABD CBAD CDBA CDAB BACD CBAD DBCA DCBA DACB DABC

推荐答案

不是您问题的直接答案，但您可以获取所有

Not a direct answer to your question but you can get all permutations (translated from java to Swift) as follow:

public extension RangeReplaceableCollection {
    func permutations() -> [SubSequence] {
        isEmpty ? [] : permutate(.init())
    }
    private func permutate(_ subSequence: SubSequence) -> [SubSequence] {
        var permutations = isEmpty ? [subSequence] : []
        indices.forEach {
            permutations += (self[..<$0] + self[$0...].dropFirst())
                .permutate(subSequence + CollectionOfOne(self[$0]))
        }
        return permutations
    }
}

---

let str = "ABCD"
print(str.permutations())   // "["ABCD", "ABDC", "ACBD", "ACDB", "ADBC", "ADCB", "BACD", "BADC", "BCAD", "BCDA", "BDAC", "BDCA", "CABD", "CADB", "CBAD", "CBDA", "CDAB", "CDBA", "DABC", "DACB", "DBAC", "DBCA", "DCAB", "DCBA"]\n"

---

对子字符串进行突变

---

Per-mutating a substring

print("ABCD".dropLast().permutations())   // ["ABC", "ACB", "BAC", "BCA", "CAB", "CBA"]\n"

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