Java 求全排列的两种方式

1 邻为互换法：

public class FullSort2 {

/*

* isExist判断j位置的字符是否已经在list[0]~list[j-1]中出现过了

* list是含重复字符的数组，i是指示当前位置的游标，j是要判断的字符的位置

*/

static boolean isExist(char a[], int k, int i) {

for (int j = k; j < i; j++) {

if (a[j] == a[i]) {

return true;

}

}

return false;

}

static void full(char[] a, int k) {

if (k == a.length - 1) {

show(a);

return;

}

for (int i = k; i < a.length; i++) {

if (isExist(a, k, i)) {

continue;

}

exch(a, i, k);

full(a, k + 1);

exch(a, i, k);

}

}

static void show(char[] a) {

for (int i = 0; i < a.length; i++)

System.out.print(a[i] + " ");

System.out.println();

}

public static void main(String[] args) {

char[] a = "ACC".toCharArray();

full(a, 0);

}

static void exch(char[] a, int i, int j) {

char swap = a[i];

a[i] = a[j];

a[j] = swap;

}

}

2 回溯法：

public class Permutation2 {

public static ArrayList<String> Permutation(String str) {

ArrayList<String> result = new ArrayList<String>();

if (str == null || str.length() == 0) {

return result;

}

char[] chars = str.toCharArray();

boolean[] used = new boolean[str.length()];

helper(chars, used, new StringBuffer(), result);

return result;

}

private static void helper(char[] chars, boolean[] used, StringBuffer sub, ArrayList<String> result) {

if (sub.length() == chars.length) {

result.add(sub.toString());

return;

}

for (int i = 0; i < used.length; i++) {

if (used[i] || i > 0 && !used[i - 1] && chars[i] == chars[i - 1]) {

continue;

}

sub.append(chars[i]);

used[i] = true;

helper(chars, used, sub, result);

sub.deleteCharAt(sub.length() - 1);

used[i] = false;

}

}

public static void main(String[] args) {

String string = "abc";

ArrayList<String> list = Permutation(string);

for (String str : list) {

System.out.println(str);

}

}

}