串的排列:

public static void main(String args[]) {
    permu(0,"ABCD");
}

static void permu(int fixed,String s) {
    char[] chr=s.toCharArray();
    if(fixed==s.length())
        System.out.println(s);
    for(int i=fixed;i<s.length();i++) {
        char c=chr[i];
        chr[i]=chr[fixed];
        chr[fixed]=c;
        permu(fixed+1,new String(chr));
    }   
}

下面是两个c#版本(仅供参考): 1. 打印所有排列 2. 返回所有排列

算法的基本要点是(可能下面的代码更直观-尽管如此，下面的代码是做什么的一些解释): -从当前索引到集合的其余部分，交换当前索引处的元素 -递归地获得下一个索引中剩余元素的排列 -恢复秩序，通过重新交换

注意:上述递归函数将从起始索引中调用。

private void PrintAllPermutations(int[] a, int index, ref int count)
        {
            if (index == (a.Length - 1))
            {
                count++;
                var s = string.Format("{0}: {1}", count, string.Join(",", a));
                Debug.WriteLine(s);
            }
            for (int i = index; i < a.Length; i++)
            {
                Utilities.swap(ref a[i], ref a[index]);
                this.PrintAllPermutations(a, index + 1, ref count);
                Utilities.swap(ref a[i], ref a[index]);
            }
        }
        private int PrintAllPermutations(int[] a)
        {
            a.ThrowIfNull("a");
            int count = 0;
            this.PrintAllPermutations(a, index:0, count: ref count);
            return count;
        }

版本2(与上面相同-但返回排列而不是打印)

private int[][] GetAllPermutations(int[] a, int index)
        {
            List<int[]> permutations = new List<int[]>();
            if (index == (a.Length - 1))
            {
                permutations.Add(a.ToArray());
            }

            for (int i = index; i < a.Length; i++)
            {
                Utilities.swap(ref a[i], ref a[index]);
                var r = this.GetAllPermutations(a, index + 1);
                permutations.AddRange(r);
                Utilities.swap(ref a[i], ref a[index]);
            }
            return permutations.ToArray();
        }
        private int[][] GetAllPermutations(int[] p)
        {
            p.ThrowIfNull("p");
            return this.GetAllPermutations(p, 0);
        }

单元测试

[TestMethod]
        public void PermutationsTests()
        {
            List<int> input = new List<int>();
            int[] output = { 0, 1, 2, 6, 24, 120 };
            for (int i = 0; i <= 5; i++)
            {
                if (i != 0)
                {
                    input.Add(i);
                }
                Debug.WriteLine("================PrintAllPermutations===================");
                int count = this.PrintAllPermutations(input.ToArray());
                Assert.IsTrue(count == output[i]);
                Debug.WriteLine("=====================GetAllPermutations=================");
                var r = this.GetAllPermutations(input.ToArray());
                Assert.IsTrue(count == r.Length);
                for (int j = 1; j <= r.Length;j++ )
                {
                    string s = string.Format("{0}: {1}", j,
                        string.Join(",", r[j - 1]));
                    Debug.WriteLine(s);
                }
                Debug.WriteLine("No.OfElements: {0}, TotalPerms: {1}", i, count);
            }
        }

使用递归。

依次尝试每个字母作为第一个字母，然后使用递归调用找到剩余字母的所有排列。 基本情况是，当输入是空字符串时，唯一的排列就是空字符串。

这是一个C解:

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>


char* addLetter(char* string, char *c) {
    char* result = malloc(sizeof(string) + 2);
    strcpy(result, string);
    strncat(result, c, 1);
    return result;
}

char* removeLetter(char* string, char *c) {
    char* result = malloc(sizeof(string));
    int j = 0;
    for (int i = 0; i < strlen(string); i++) {
        if (string[i] != *c) {
            result[j++] = string[i];
        }
    }
    result[j] = '\0';

    return result;
}

void makeAnagram(char *anagram, char *letters) {

    if (*letters == '\0') {
        printf("%s\n", anagram);
        return;
    }

    char *c = letters;
    while (*c != '\0') {
        makeAnagram(addLetter(anagram, c),
                    removeLetter(letters, c));
        c++;
    }

}

int main() {

    makeAnagram("", "computer");

    return 0;
}

基于Heap算法的我的实现:

import java.util.ArrayList;
import java.util.List;

public class PermutationString {
public static List<String> permute(char[] str, int n) {
    List<String> permutations = new ArrayList<>();
    if (n == 1) {
        permutations.add(new String(str));
    }
    else {
        for (int i = 0; i < n; i++) {
            permutations.addAll(permute(str, n-1));
            if (n % 2 == 0) {
                swap(str, i, n-1);
            }
            else {
                swap(str, 0, n-1);
            }
        }
    }
    return permutations;
}


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

public static void main(String[] args) {

    List<String> permutations = permute("abcdefgh".toCharArray(), 8);

    System.out.println(permutations);

}
}

时间复杂度为O(n!* n)， O(n)为空间复杂度。

public static void permutation(String str) { 
    permutation("", str); 
}

private static void permutation(String prefix, String str) {
    int n = str.length();
    if (n == 0) System.out.println(prefix);
    else {
        for (int i = 0; i < n; i++)
            permutation(prefix + str.charAt(i), str.substring(0, i) + str.substring(i+1, n));
    }
}

(通过Java编程入门)