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
如果有人想要生成排列来做一些事情,而不是通过void方法打印它们:
static List<int[]> permutations(int n) {
class Perm {
private final List<int[]> permutations = new ArrayList<>();
private void perm(int[] array, int step) {
if (step == 1) permutations.add(array.clone());
else for (int i = 0; i < step; i++) {
perm(array, step - 1);
int j = (step % 2 == 0) ? i : 0;
swap(array, step - 1, j);
}
}
private void swap(int[] array, int i, int j) {
int buffer = array[i];
array[i] = array[j];
array[j] = buffer;
}
}
int[] nVector = new int[n];
for (int i = 0; i < n; i++) nVector [i] = i;
Perm perm = new Perm();
perm.perm(nVector, n);
return perm.permutations;
}
其他回答
简单的递归c++实现如下所示:
#include <iostream>
void generatePermutations(std::string &sequence, int index){
if(index == sequence.size()){
std::cout << sequence << "\n";
} else{
generatePermutations(sequence, index + 1);
for(int i = index + 1 ; i < sequence.size() ; ++i){
std::swap(sequence[index], sequence[i]);
generatePermutations(sequence, index + 1);
std::swap(sequence[index], sequence[i]);
}
}
}
int main(int argc, char const *argv[])
{
std::string str = "abc";
generatePermutations(str, 0);
return 0;
}
输出:
abc
acb
bac
bca
cba
cab
更新
如果想要存储结果,可以将vector作为函数调用的第三个参数传递。此外,如果您只想要唯一的排列,您可以使用集合。
#include <iostream>
#include <vector>
#include <set>
void generatePermutations(std::string &sequence, int index, std::vector <std::string> &v){
if(index == sequence.size()){
//std::cout << sequence << "\n";
v.push_back(sequence);
} else{
generatePermutations(sequence, index + 1, v);
for(int i = index + 1 ; i < sequence.size() ; ++i){
std::swap(sequence[index], sequence[i]);
generatePermutations(sequence, index + 1, v);
std::swap(sequence[index], sequence[i]);
}
}
}
int main(int argc, char const *argv[])
{
std::string str = "112";
std::vector <std::string> permutations;
generatePermutations(str, 0, permutations);
std::cout << "Number of permutations " << permutations.size() << "\n";
for(const std::string &s : permutations){
std::cout << s << "\n";
}
std::set <std::string> uniquePermutations(permutations.begin(), permutations.end());
std::cout << "Number of unique permutations " << uniquePermutations.size() << "\n";
for(const std::string &s : uniquePermutations){
std::cout << s << "\n";
}
return 0;
}
输出:
Number of permutations 6
112
121
112
121
211
211
Number of unique permutations 3
112
121
211
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','')
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
}
}
使用Es6的字符串排列
使用reduce()方法
Const排列= STR => { If (str.length <= 2) 返回str.length === 2 ?[str, str[1] + str[0]]: [str]; 返回str .split (") .reduce ( (acc, letter, index) => acc.concat(排列(str。Slice (0, index) + str.slice(index + 1))。Map (val =>字母+ val)), [] ); }; console.log(排列(STR));
使用递归的简单python解决方案。
def get_permutations(string):
# base case
if len(string) <= 1:
return set([string])
all_chars_except_last = string[:-1]
last_char = string[-1]
# recursive call: get all possible permutations for all chars except last
permutations_of_all_chars_except_last = get_permutations(all_chars_except_last)
# put the last char in all possible positions for each of the above permutations
permutations = set()
for permutation_of_all_chars_except_last in permutations_of_all_chars_except_last:
for position in range(len(all_chars_except_last) + 1):
permutation = permutation_of_all_chars_except_last[:position] + last_char + permutation_of_all_chars_except_last[position:]
permutations.add(permutation)
return permutations
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