java - 两个字符的构造模式(例如 : "aabba" ,"aba")

Question

在给定两个字符的出现频率(例如:x = 5)的情况下，建议一种使用两个字符(例如:“aabba”、“aba”)的模式构造 String 的有效方法, y = 4).问题是任何字符都不应重复超过两次。

示例测试用例:

|  X  |  Y  | Output     |
| --- | --- | ---------- |
|  3  |  2  | aabab      |
|  2  |  1  | aab        |
|  4  |  6  | bbabbababa |
|  4  |  4  | abababab   |

我写了一个贪心的方法如下，

public static String getPatternStr(int x, int y){
    String result = "";
    List<String> list = new ArrayList<>();
    int secondIterationIndex = 0;

    int moreCharCount = (x > y)? x : y;
    int lessCharCount = (x < y)? x : y;


    String moreChar = (x > y)?"a":"b";
    String lessChar = (x < y)?"a":"b";

    if(x == y){
        moreCharCount = lessCharCount = x;
        moreChar = "a";
        lessChar = "b";
    }

    for(int i = 1; i <= (x+y); i++){
        if(lessCharCount > 0){
            if(i%2 == 1){
                list.add(moreChar);
                moreCharCount--;
            }
            else{
                list.add(lessChar);
                lessCharCount--;
            }
        }else{
            list.add(secondIterationIndex,moreChar);
            secondIterationIndex += 3;
            if(secondIterationIndex > list.size()){
                secondIterationIndex = list.size()-1; 
            }
        }
        //System.out.println(list);
    }
    for(String e: list){
        result += e;
    }
    return result;
}

这种方法看起来很繁琐，也不优雅。请问有什么更好更有效的方法吗？

编辑:我坚信有一种使用这些(x 和 y)数字的方法，我们甚至可以在循环开始之前计算双 a、双 b、单 a 和单 b 的数量。只有部分工作逻辑。

public static String patternStr(int x, int y){
    String result = "";

    int large = (x > y)? x : y;
    int small = (x < y)? x : y;


    int pairsOfLarge = large/2;
    int largeParts = pairsOfLarge + large%2;

    int minSmallParts = (largeParts>1)? (largeParts-1):1;
    int pairsOfSmall = small - minSmallParts;
    int smallParts = pairsOfSmall + (small - (pairsOfSmall*2));
    // System.out.println("minSmallParts="+minSmallParts+" pairsOfSmall="+pairsOfSmall+" smallParts="+smallParts);

    String odd = (large == x)?"a":"b";
    String even = (small == x)?"a":"b";

    int i = 1;
    while((largeParts + smallParts) > 0){
        if(i%2 > 0){
            if(pairsOfLarge > 0){
                result += odd + odd;
                pairsOfLarge--;
            }else{
                result += odd;
            }
            largeParts--;
        }else{
            if(pairsOfSmall > 0){
                result += even + even;
                pairsOfSmall--;
            }else{
                result += even;
            }
            smallParts--;
        }
        i++;
    }
    return result;
}

Answer 1

这对我来说更优雅(算法的工作是相似的):

static String getPatternStr(int x, int y){
    int lessCharCount = Math.min(x,y);
    int moreCharCount = Math.max(x,y);
    String moreChar = "a";
    String lessChar = "b";
    if(lessCharCount < (moreCharCount + 1) / 2 - 1)
      return "";
    if(lessCharCount == 1 && moreCharCount == 1)
      return moreChar + lessChar;
    LinkedList<String> result = new LinkedList<>(Arrays.asList(moreChar.repeat(moreCharCount).split("")));
    for(int i = 2; lessCharCount > 0; i = (i + 3) % (++moreCharCount)){
      result.add(i,lessChar);
      lessCharCount--;
    }
    return result.stream().collect(Collectors.joining(""));
  }

没有太大的效率提升，只是使用了LinkedList而不是ArraysList，前者更适合插入操作。