为排列和组合添加更详细的NcK/NcR

public static void combinationNcK(List<String> inputList, String prefix, int chooseCount, List<String> resultList) {
    if (chooseCount == 0)
        resultList.add(prefix);
    else {
        for (int i = 0; i < inputList.size(); i++)
            combinationNcK(inputList.subList(i + 1, inputList.size()), prefix + "," + inputList.get(i), chooseCount - 1, resultList);

        // Finally print once all combinations are done
        if (prefix.equalsIgnoreCase("")) {
            resultList.stream().map(str -> str.substring(1)).forEach(System.out::println);
        }
    }
}

public static void permNcK(List<String> inputList, int chooseCount, List<String> resultList) {
    for (int count = 0; count < inputList.size(); count++) {
        permNcK(inputList, "", chooseCount, resultList);
        resultList = new ArrayList<String>();
        Collections.rotate(inputList, 1);
        System.out.println("-------------------------");
    }

}

public static void permNcK(List<String> inputList, String prefix, int chooseCount, List<String> resultList) {
    if (chooseCount == 0)
        resultList.add(prefix);
    else {
        for (int i = 0; i < inputList.size(); i++)
            combinationNcK(inputList.subList(i + 1, inputList.size()), prefix + "," + inputList.get(i), chooseCount - 1, resultList);

        // Finally print once all combinations are done
        if (prefix.equalsIgnoreCase("")) {
            resultList.stream().map(str -> str.substring(1)).forEach(System.out::println);
        }
    }
}

public static void main(String[] args) {
    List<String> positions = Arrays.asList(new String[] { "1", "2", "3", "4", "5", "6", "7", "8" });
    List<String> resultList = new ArrayList<String>();
    //combinationNcK(positions, "", 3, resultList);

    permNcK(positions, 3, resultList);

}

使用递归。

当输入是空字符串时，唯一的排列就是空字符串。尝试将字符串中的每个字母作为第一个字母，然后使用递归调用找到其余字母的所有排列。

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

class Permutation {
    private static List<String> permutation(String prefix, String str) {
        List<String> permutations = new ArrayList<>();
        int n = str.length();
        if (n == 0) {
            permutations.add(prefix);
        } else {
            for (int i = 0; i < n; i++) {
                permutations.addAll(permutation(prefix + str.charAt(i), str.substring(i + 1, n) + str.substring(0, i)));
            }
        }
        return permutations;
    }

    public static void main(String[] args) {
        List<String> perms = permutation("", "abcd");

        String[] array = new String[perms.size()];
        for (int i = 0; i < perms.size(); i++) {
            array[i] = perms.get(i);
        }

        int x = array.length;

        for (final String anArray : array) {
            System.out.println(anArray);
        }
    }
}

这里有一个优雅的，非递归的O(n!)解:

public static StringBuilder[] permutations(String s) {
        if (s.length() == 0)
            return null;
        int length = fact(s.length());
        StringBuilder[] sb = new StringBuilder[length];
        for (int i = 0; i < length; i++) {
            sb[i] = new StringBuilder();
        }
        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            int times = length / (i + 1);
            for (int j = 0; j < times; j++) {
                for (int k = 0; k < length / times; k++) {
                    sb[j * length / times + k].insert(k, ch);
                }
            }
        }
        return sb;
    }

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));
    }   
}

python实现

def getPermutation(s, prefix=''):
        if len(s) == 0:
                print prefix
        for i in range(len(s)):
                getPermutation(s[0:i]+s[i+1:len(s)],prefix+s[i] )



getPermutation('abcd','')

作为Python生成器，带有现代类型提示:

from typing import Iterator


def permutations(string: str, prefix: str = '') -> Iterator[str]:
    if len(string) == 0:
        yield prefix
    for i, character in enumerate(string):
        yield from permutations(string[:i] + string[i + 1:], prefix + character)


for p in permutations('abcd'):
    print(p)