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
我定义了左右两个字符串。一开始,左边是输入字符串,右边是“”。我递归地从左边选择所有可能的字符,并将其添加到右边的末尾。然后,在left-charAt(I)和right+charAt(I)上调用递归函数。我定义了一个类来跟踪生成的排列。
import java.util.HashSet;
import java.util.Set;
public class FindPermutations {
static class Permutations {
Set<String> permutations = new HashSet<>();
}
/**
* Building all the permutations by adding chars of left to right one by one.
*
* @param left The left string
* @param right The right string
* @param permutations The permutations
*/
private void findPermutations(String left, String right, Permutations permutations) {
int n = left.length();
if (n == 0) {
permutations.permutations.add(right);
}
for (int i = 0; i < n; i++) {
findPermutations(left.substring(0, i) + left.substring(i + 1, n), right + left.charAt(i), permutations);
}
}
/**
* Gets all the permutations of a string s.
*
* @param s The input string
* @return all the permutations of a string s
*/
public Permutations getPermutations(String s) {
Permutations permutations = new Permutations();
findPermutations(s, "", permutations);
return permutations;
}
public static void main(String[] args) {
FindPermutations findPermutations = new FindPermutations();
String s = "ABC";
Permutations permutations = findPermutations.getPermutations(s);
printPermutations(permutations);
}
private static void printPermutations(Permutations permutations) {
for (String p : permutations.permutations) {
System.out.println(p);
}
}
}
我希望这能有所帮助。
其他回答
使用递归。
依次尝试每个字母作为第一个字母,然后使用递归调用找到剩余字母的所有排列。 基本情况是,当输入是空字符串时,唯一的排列就是空字符串。
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));
Java中一个非常基本的解决方案是使用递归+设置(以避免重复),如果你想存储和返回解决方案字符串:
public static Set<String> generatePerm(String input)
{
Set<String> set = new HashSet<String>();
if (input == "")
return set;
Character a = input.charAt(0);
if (input.length() > 1)
{
input = input.substring(1);
Set<String> permSet = generatePerm(input);
for (String x : permSet)
{
for (int i = 0; i <= x.length(); i++)
{
set.add(x.substring(0, i) + a + x.substring(i));
}
}
}
else
{
set.add(a + "");
}
return set;
}
//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);
}
}
}
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)
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