swift：比較2個字符數組，從每個字符中刪除一個值

Question

這是我正在嘗試做的。我正在製作一個遊戲，讓對手猜對方的話，每個猜測都有直接和間接的點擊。請參閱星號，瞭解我遇到問題的位置。

var directSum = 0 
var indirectSum = 0 
var words = ["canon", "cnams"] 
var secretWord = Array(words[0].characters) 
var guessWord = Array(words[1].characters) 
var secretWordInd = [Character]() 
var guessWordInd = [Character]() 

let dir1 = secretWord[0] == guessWord[0] 
let dir2 = secretWord[1] == guessWord[1] 
let dir3 = secretWord[2] == guessWord[2] 
let dir4 = secretWord[3] == guessWord[3] 
let dir5 = secretWord[4] == guessWord[4] 


if dir1 && dir2 && dir3 && dir4 && dir5 { 
    print ("you won!") 
} 
    if dir1 { 
     directSum += 1 
    } else { 
     secretWordInd.append(secretWord[0]) 
     guessWordInd.append(guessWord[0]) 
} 
    if dir2 { 
     directSum += 1 
    } else { 
     secretWordInd.append(secretWord[1]) 
     guessWordInd.append(guessWord[1]) 
} 
    if dir3 { 
     directSum += 1 
    } else { 
     secretWordInd.append(secretWord[2]) 
     guessWordInd.append(guessWord[2]) 
} 
    if dir4 { 
     directSum += 1 
    } else { 
     secretWordInd.append(secretWord[3]) 
     guessWordInd.append(guessWord[3]) 
} 
    if dir5 { 
     directSum += 1 
    } else { 
     secretWordInd.append(secretWord[4]) 
     guessWordInd.append(guessWord[4]) 
**} 

for var secretLetter in secretWordInd { 
    for var guessLetter in guessWordInd{ 
     if secretLetter == guessLetter { 
      secretWordInd.remove(at:secretWordInd.indexOf(secretLetter)) 
      guessWordInd.remove(at:guessWordInd.indexOf(guessLetter)) 
      indirectSum += 1 
      } 
     } 
    }** 
var score = [directSum, indirectSum] 

什麼，我需要做的是計算每一個有一個在數組一個字符時間：SecretWordInd在陣列相匹配的字符：guessWordInd，只刪除這兩個字符（每個字符串），並indirectSum + = 1每次發生這種情況。例如，如果一箇中有2個「a」，另一箇中有1個，則該功能需要從每個中刪除1個a。 這意味着directSum函數的輸出，在這種情況下，indirectSum應該是[1,2]，因爲從猜測詞中的一個「n」只有一個間接命中。 這是是什麼使我的功能變得複雜。這兩個數組中的值數量不會保持不變。我無法弄清楚如何使用一種方法來做到這一點（.contains僅適用於我認爲的字符串）。非常感謝您的幫助！謝謝！

Answer 1

你應該看看Array中，在文檔中設置和字符串數據類型，它們提供了大量的函數來執行那種計算和轉換。

例如：

var secretWord = "canon" 
var guessWord = "cnams" 

// identify shared letters using sets 
let sharedLetters = Set(secretWord.characters).intersection(Set(guessWord.characters)) 

// count number of shared letters present in both words 
let sharedCounts = sharedLetters.map{ c  in (c, secretWord.characters, guessWord.characters) } 
            .map{ (c,s,g) in (c, s.filter{$0==c}, g.filter{$0==c}) } 
            .map{ (c,s,g) in (c, min(s.count,g.count)) } 

// count same letters at same position and shared 
// (if you're making a mastermind game, you should subtract directCount from indirectCount)         
let directCount = zip(secretWord.characters,guessWord.characters).filter{$0==$1}.count 
let indirectCount = sharedCounts.map{$0.1}.reduce(0,+)         


// remove shared letters (for count) from both strings 
sharedCounts.flatMap{ Array(repeating:$0, count:$1) } 
      .forEach{ secretWord.remove(at:secretWord.characters.index(of:$0)!) } 
sharedCounts.flatMap{ Array(repeating:$0, count:$1) } 
      .forEach{ guessWord.remove(at:guessWord.characters.index(of:$0)!) } 


sharedLetters // {"n", "a", "c"} 
sharedCounts // ("n", 1), ("a",1), ("c", 1)] 
directCount // 1 
indirectCount // 3  
secretWord  // "on" 
guessWord  // "ms" 

Answer 2

正如其他人所說，這不是功課的地方;）

但這裏有一些指針，以什麼可能是有用的：

的indexOf：返回元素的索引你正在尋找，無如果不設定

刪除：從陣列