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
我们可以用阶乘来计算有多少字符串以某个字母开头。
示例:取输入abcd。(3!) == 6个字符串将以abcd中的每个字母开头。
static public int facts(int x){
int sum = 1;
for (int i = 1; i < x; i++) {
sum *= (i+1);
}
return sum;
}
public static void permutation(String str) {
char[] str2 = str.toCharArray();
int n = str2.length;
int permutation = 0;
if (n == 1) {
System.out.println(str2[0]);
} else if (n == 2) {
System.out.println(str2[0] + "" + str2[1]);
System.out.println(str2[1] + "" + str2[0]);
} else {
for (int i = 0; i < n; i++) {
if (true) {
char[] str3 = str.toCharArray();
char temp = str3[i];
str3[i] = str3[0];
str3[0] = temp;
str2 = str3;
}
for (int j = 1, count = 0; count < facts(n-1); j++, count++) {
if (j != n-1) {
char temp1 = str2[j+1];
str2[j+1] = str2[j];
str2[j] = temp1;
} else {
char temp1 = str2[n-1];
str2[n-1] = str2[1];
str2[1] = temp1;
j = 1;
} // end of else block
permutation++;
System.out.print("permutation " + permutation + " is -> ");
for (int k = 0; k < n; k++) {
System.out.print(str2[k]);
} // end of loop k
System.out.println();
} // end of loop j
} // end of loop i
}
}
其他回答
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)
在这里和其他论坛给出的所有解决方案中,我最喜欢Mark Byers。这个描述实际上让我自己思考并编写了代码。 可惜我不能投票支持他的解决方案,因为我是新手。 无论如何,这是我对他的描述的实现
public class PermTest {
public static void main(String[] args) throws Exception {
String str = "abcdef";
StringBuffer strBuf = new StringBuffer(str);
doPerm(strBuf,0);
}
private static void doPerm(StringBuffer str, int index){
if(index == str.length())
System.out.println(str);
else { //recursively solve this by placing all other chars at current first pos
doPerm(str, index+1);
for (int i = index+1; i < str.length(); i++) {//start swapping all other chars with current first char
swap(str,index, i);
doPerm(str, index+1);
swap(str,i, index);//restore back my string buffer
}
}
}
private static void swap(StringBuffer str, int pos1, int pos2){
char t1 = str.charAt(pos1);
str.setCharAt(pos1, str.charAt(pos2));
str.setCharAt(pos2, t1);
}
}
我更喜欢这个解决方案,而不是第一个解决方案,因为这个解决方案使用StringBuffer。我不会说我的解决方案没有创建任何临时字符串(它实际上在system.out.println中创建,其中调用StringBuffer的toString())。但我只是觉得这比第一个解决方案好太多的字符串字面值被创建。可能有些性能人员可以根据“内存”来评估这一点(对于“时间”来说,由于额外的“交换”,它已经滞后了)
这是一个更快的解决方案,因为它不受字符串连接计算复杂度O(n^2)的影响。另一方面它是无循环的,完全递归的
public static void main(String[] args) {
permutation("ABCDEFGHIJKLMNOPQRSTUVWXYZ");
}
private static void permutation(String str) {
char[] stringArray = str.toCharArray();
printPermutation(stringArray, 0, stringArray.length, 0, 1);
}
private static void printPermutation(char[] string, int loopCounter, int length, int indexFrom, int indexTo) {
// Stop condition
if (loopCounter == length)
return;
/*
When reaching the end of the array:
1- Reset loop indices.
2- Increase length counter.
*/
if (indexTo == length) {
indexFrom = 0;
indexTo = 1;
++loopCounter;
}
// Print.
System.out.println(string);
// Swap from / to indices.
char temp = string[indexFrom];
string[indexFrom] = string[indexTo];
string[indexTo] = temp;
// Go for next iteration.
printPermutation(string, loopCounter, length, ++indexFrom, ++indexTo);
}
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','')
让我们以输入abc为例。
从集合(["c"])中的最后一个元素(c)开始,然后将最后第二个元素(b)添加到它的前面,末尾和中间的每个可能位置,使其["bc", "cb"],然后以同样的方式将后面的下一个元素(a)添加到集合中的每个字符串中,使其:
"a" + "bc" = ["abc", "bac", "bca"] and "a" + "cb" = ["acb" ,"cab", "cba"]
因此整个排列:
["abc", "bac", "bca","acb" ,"cab", "cba"]
代码:
public class Test
{
static Set<String> permutations;
static Set<String> result = new HashSet<String>();
public static Set<String> permutation(String string) {
permutations = new HashSet<String>();
int n = string.length();
for (int i = n - 1; i >= 0; i--)
{
shuffle(string.charAt(i));
}
return permutations;
}
private static void shuffle(char c) {
if (permutations.size() == 0) {
permutations.add(String.valueOf(c));
} else {
Iterator<String> it = permutations.iterator();
for (int i = 0; i < permutations.size(); i++) {
String temp1;
for (; it.hasNext();) {
temp1 = it.next();
for (int k = 0; k < temp1.length() + 1; k += 1) {
StringBuilder sb = new StringBuilder(temp1);
sb.insert(k, c);
result.add(sb.toString());
}
}
}
permutations = result;
//'result' has to be refreshed so that in next run it doesn't contain stale values.
result = new HashSet<String>();
}
}
public static void main(String[] args) {
Set<String> result = permutation("abc");
System.out.println("\nThere are total of " + result.size() + " permutations:");
Iterator<String> it = result.iterator();
while (it.hasNext()) {
System.out.println(it.next());
}
}
}
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