找出弦的所有排列的优雅方法是什么。例如,ba的排列,将是ba和ab,但更长的字符串,如abcdefgh?是否有Java实现示例?
当前回答
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','')
其他回答
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)
//循环'整个字符数组,并保持'i'作为你的排列的基础,并像你交换[ab, ba]一样继续寻找组合
public class Permutation {
//Act as a queue
private List<Character> list;
//To remove the duplicates
private Set<String> set = new HashSet<String>();
public Permutation(String s) {
list = new LinkedList<Character>();
int len = s.length();
for(int i = 0; i < len; i++) {
list.add(s.charAt(i));
}
}
public List<String> getStack(Character c, List<Character> list) {
LinkedList<String> stack = new LinkedList<String>();
stack.add(""+c);
for(Character ch: list) {
stack.add(""+ch);
}
return stack;
}
public String printCombination(String s1, String s2) {
//S1 will be a single character
StringBuilder sb = new StringBuilder();
String[] strArr = s2.split(",");
for(String s: strArr) {
sb.append(s).append(s1);
sb.append(",");
}
for(String s: strArr) {
sb.append(s1).append(s);
sb.append(",");
}
return sb.toString();
}
public void printPerumtation() {
int cnt = list.size();
for(int i = 0; i < cnt; i++) {
Character c = list.get(0);
list.remove(0);
List<String> stack = getStack(c, list);
while(stack.size() > 1) {
//Remove the top two elements
String s2 = stack.remove(stack.size() - 1);
String s1 = stack.remove(stack.size() - 1);
String comS = printCombination(s1, s2);
stack.add(comS);
}
String[] perms = (stack.remove(0)).split(",");
for(String perm: perms) {
set.add(perm);
}
list.add(c);
}
for(String s: set) {
System.out.println(s);
}
}
}
倒计时Quickperm算法的通用实现,表示#1(可伸缩,非递归)。
/**
* Generate permutations based on the
* Countdown <a href="http://quickperm.org/">Quickperm algorithm</>.
*/
public static <T> List<List<T>> generatePermutations(List<T> list) {
List<T> in = new ArrayList<>(list);
List<List<T>> out = new ArrayList<>(factorial(list.size()));
int n = list.size();
int[] p = new int[n +1];
for (int i = 0; i < p.length; i ++) {
p[i] = i;
}
int i = 0;
while (i < n) {
p[i]--;
int j = 0;
if (i % 2 != 0) { // odd?
j = p[i];
}
// swap
T iTmp = in.get(i);
in.set(i, in.get(j));
in.set(j, iTmp);
i = 1;
while (p[i] == 0){
p[i] = i;
i++;
}
out.add(new ArrayList<>(in));
}
return out;
}
private static int factorial(int num) {
int count = num;
while (num != 1) {
count *= --num;
}
return count;
}
它需要list,因为泛型不能很好地使用数组。
另一种简单的方法是遍历字符串,选择尚未使用的字符并将其放入缓冲区,继续循环,直到缓冲区大小等于字符串长度。我更喜欢这个回溯跟踪解决方案,因为:
容易理解 容易避免重复 输出是排序的
下面是java代码:
List<String> permute(String str) {
if (str == null) {
return null;
}
char[] chars = str.toCharArray();
boolean[] used = new boolean[chars.length];
List<String> res = new ArrayList<String>();
StringBuilder sb = new StringBuilder();
Arrays.sort(chars);
helper(chars, used, sb, res);
return res;
}
void helper(char[] chars, boolean[] used, StringBuilder sb, List<String> res) {
if (sb.length() == chars.length) {
res.add(sb.toString());
return;
}
for (int i = 0; i < chars.length; i++) {
// avoid duplicates
if (i > 0 && chars[i] == chars[i - 1] && !used[i - 1]) {
continue;
}
// pick the character that has not used yet
if (!used[i]) {
used[i] = true;
sb.append(chars[i]);
helper(chars, used, sb, res);
// back tracking
sb.deleteCharAt(sb.length() - 1);
used[i] = false;
}
}
}
str输入:1231
输出列表:{1123,1132,1213,1231,1312,1321,2113,2131,2311,3112,3121,3211}
注意,输出是排序的,没有重复的结果。
基于Mark Byers的回答,我的python实现:
def permutations(string):
if len(string) == 1:
return [string]
permutations=[]
for i in range(len(string)):
for perm in permutations(string[:i]+string[i+1:]):
permutations.append(string[i] + perm)
return permutations
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