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
//插入每个字符到数组列表中
static ArrayList al = new ArrayList();
private static void findPermutation (String str){
for (int k = 0; k < str.length(); k++) {
addOneChar(str.charAt(k));
}
}
//insert one char into ArrayList
private static void addOneChar(char ch){
String lastPerStr;
String tempStr;
ArrayList locAl = new ArrayList();
for (int i = 0; i < al.size(); i ++ ){
lastPerStr = al.get(i).toString();
//System.out.println("lastPerStr: " + lastPerStr);
for (int j = 0; j <= lastPerStr.length(); j++) {
tempStr = lastPerStr.substring(0,j) + ch +
lastPerStr.substring(j, lastPerStr.length());
locAl.add(tempStr);
//System.out.println("tempStr: " + tempStr);
}
}
if(al.isEmpty()){
al.add(ch);
} else {
al.clear();
al = locAl;
}
}
private static void printArrayList(ArrayList al){
for (int i = 0; i < al.size(); i++) {
System.out.print(al.get(i) + " ");
}
}
其他回答
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
}
}
这里有一个优雅的,非递归的O(n!)解:
public static StringBuilder[] permutations(String s) {
if (s.length() == 0)
return null;
int length = fact(s.length());
StringBuilder[] sb = new StringBuilder[length];
for (int i = 0; i < length; i++) {
sb[i] = new StringBuilder();
}
for (int i = 0; i < s.length(); i++) {
char ch = s.charAt(i);
int times = length / (i + 1);
for (int j = 0; j < times; j++) {
for (int k = 0; k < length / times; k++) {
sb[j * length / times + k].insert(k, ch);
}
}
}
return sb;
}
下面是两个c#版本(仅供参考): 1. 打印所有排列 2. 返回所有排列
算法的基本要点是(可能下面的代码更直观-尽管如此,下面的代码是做什么的一些解释): -从当前索引到集合的其余部分,交换当前索引处的元素 -递归地获得下一个索引中剩余元素的排列 -恢复秩序,通过重新交换
注意:上述递归函数将从起始索引中调用。
private void PrintAllPermutations(int[] a, int index, ref int count)
{
if (index == (a.Length - 1))
{
count++;
var s = string.Format("{0}: {1}", count, string.Join(",", a));
Debug.WriteLine(s);
}
for (int i = index; i < a.Length; i++)
{
Utilities.swap(ref a[i], ref a[index]);
this.PrintAllPermutations(a, index + 1, ref count);
Utilities.swap(ref a[i], ref a[index]);
}
}
private int PrintAllPermutations(int[] a)
{
a.ThrowIfNull("a");
int count = 0;
this.PrintAllPermutations(a, index:0, count: ref count);
return count;
}
版本2(与上面相同-但返回排列而不是打印)
private int[][] GetAllPermutations(int[] a, int index)
{
List<int[]> permutations = new List<int[]>();
if (index == (a.Length - 1))
{
permutations.Add(a.ToArray());
}
for (int i = index; i < a.Length; i++)
{
Utilities.swap(ref a[i], ref a[index]);
var r = this.GetAllPermutations(a, index + 1);
permutations.AddRange(r);
Utilities.swap(ref a[i], ref a[index]);
}
return permutations.ToArray();
}
private int[][] GetAllPermutations(int[] p)
{
p.ThrowIfNull("p");
return this.GetAllPermutations(p, 0);
}
单元测试
[TestMethod]
public void PermutationsTests()
{
List<int> input = new List<int>();
int[] output = { 0, 1, 2, 6, 24, 120 };
for (int i = 0; i <= 5; i++)
{
if (i != 0)
{
input.Add(i);
}
Debug.WriteLine("================PrintAllPermutations===================");
int count = this.PrintAllPermutations(input.ToArray());
Assert.IsTrue(count == output[i]);
Debug.WriteLine("=====================GetAllPermutations=================");
var r = this.GetAllPermutations(input.ToArray());
Assert.IsTrue(count == r.Length);
for (int j = 1; j <= r.Length;j++ )
{
string s = string.Format("{0}: {1}", j,
string.Join(",", r[j - 1]));
Debug.WriteLine(s);
}
Debug.WriteLine("No.OfElements: {0}, TotalPerms: {1}", i, count);
}
}
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));
简单的解决方案,利用swift语言的特点,数组是值类型。
func permutation(chrs: [String], arr: [String], result: inout [[String]]) {
if arr.count == chrs.count {
result.append(arr)
return
}
for chr in chrs {
var arr = arr
if !arr.contains(chr) {
arr.append(chr)
permutation(chrs: chrs, arr: arr, result: &result)
}
}
}
func test() {
var result = [[String]]()
let chrs = ["a", "b", "c", "d"]
permutation(chrs: chrs, arr: [], result: &result)
}
复杂度O(n * n!)
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