这是一个示例实现。本质上，它需要一个String并遍历每个字符，然后将该字符放在最前面。然后，它在其余字符上递归。该结构消除了重复字母的问题，因为递归的输入已删除了您已经使用的字符。
我还将结果存储在一个集合中，以消除语义上的对等。输入“ aab”可以切换char 0和char 1，但仍为“
aab”。我使用TreeSet保留顺序以便更轻松地验证输出，但是HashSet会更快。
public static Set<String> permute(String chars) { // Use sets to eliminate semantic duplicates (aab is still aab even if you switch the two 'a's) // Switch to HashSet for better performance Set<String> set = new TreeSet<String>(); // Termination condition: only 1 permutation for a string of length 1 if (chars.length() == 1) { set.add(chars); } else { // Give each character a chance to be the first in the permuted string for (int i=0; i<chars.length(); i++) { // Remove the character at index i from the string String pre = chars.substring(0, i); String post = chars.substring(i+1); String remaining = pre+post; // Recurse to find all the permutations of the remaining chars for (String permutation : permute(remaining)) { // Concatenate the first character with the permutations of the remaining chars set.add(chars.charAt(i) + permutation); } } } return set; }示例运行：
public static void main(String[] args) { for (String s : CharPermuter.permute("abca")) { System.out.println(s); } }产生：
aabcaacbabacabcaacabacbabaacbacabcaacaabcabacbaa
