使用迭代对字符串进行排列

Question

我正在尝试查找给定字符串的排列，但是我想使用迭代。我在网上找到了递归解决方案，但我确实理解它，但是将其转换为迭代解决方案实际上是行不通的。下面附上我的代码。我非常感谢您的帮助：

public static void combString(String s) {
    char[] a = new char[s.length()];
    //String temp = "";
    for(int i = 0; i < s.length(); i++) {
        a[i] = s.charAt(i);
    }
    for(int i = 0; i < s.length(); i++) {
        String temp = "" + a[i];

        for(int j = 0; j < s.length();j++) {
            //int k = j;
            if(i != j) {
                System.out.println(j);
                temp += s.substring(0,j) + s.substring(j+1,s.length());
            }               
        }
        System.out.println(temp);
    }
}

Answer 1

在我的相关问题评论之后，这是一个Java实现，可以使用Counting QuickPerm
Algorithm
来完成您想要的事情：

public static void combString(String s) {
    // Print initial string, as only the alterations will be printed later
    System.out.println(s);   
    char[] a = s.toCharArray();
    int n = a.length;
    int[] p = new int[n];  // Weight index control array initially all zeros. Of course, same size of the char array.
    int i = 1; //Upper bound index. i.e: if string is "abc" then index i could be at "c"
    while (i < n) {
        if (p[i] < i) { //if the weight index is bigger or the same it means that we have already switched between these i,j (one iteration before).
            int j = ((i % 2) == 0) ? 0 : p[i];//Lower bound index. i.e: if string is "abc" then j index will always be 0.
            swap(a, i, j);
            // Print current
            System.out.println(join(a));
            p[i]++; //Adding 1 to the specific weight that relates to the char array.
            i = 1; //if i was 2 (for example), after the swap we now need to swap for i=1
        }
        else { 
            p[i] = 0;//Weight index will be zero because one iteration before, it was 1 (for example) to indicate that char array a[i] swapped.
            i++;//i index will have the option to go forward in the char array for "longer swaps"
        }
    }
}

private static String join(char[] a) {
    StringBuilder builder = new StringBuilder();
    builder.append(a);
    return builder.toString();
}

private static void swap(char[] a, int i, int j) {
    char temp = a[i];
    a[i] = a[j];
    a[j] = temp;
}