求一个字符串的全部排列，网上找到的全部方法要么是递归，要么是三层for循环。 java
我这个方法不须要递归，只要两层for循环，思路很是独特。缺点是字符串长度较长时执行时间可能要几万年。但毕竟是本身的原创思路，且网上没有这个算法，姑且记录下来。算法
import java.util.ArrayList;
import java.util.List;
import java.util.Random;
public class Main {
private List<String> elements = new ArrayList<>();
private int length;
public Main(String input) {
length = input.length();
for (int i = 0; i < length; i++) {
elements.add("");
}
for (int i = 0; i < length; i++) {
String element = input.substring(i, i + 1);
elements.set(i, element);
}
}
private List<String> permutate() {
List<String> permutations = new ArrayList<>();
for (int p = 0; p < Short.MAX_VALUE; p++) { // 字符串较长时须要改为一个很大的数
StringBuilder randomSb = new StringBuilder();
for (int i = 0; i < length; i++) {
int r = new Random().nextInt(length);
if (!randomSb.toString().contains(elements.get(r))) {
randomSb.append(elements.get(r));
}
}
if (!permutations.contains(randomSb.toString()) && randomSb.toString().length() == length) {
permutations.add(randomSb.toString());
}
}
return permutations;
}
public static void main(String[] args) {
Main main = new Main("ABCDE");
List<String> result = main.permutate();
for (String s : result) {
System.out.println(s);
}
System.out.println("Number of permutations is " + result.size());
}
}
执行结果：json
BEDCA EBADC ADBCE DCEBA ECDAB DEACB DACEB CBDAE EDACB DEBCA ACBDE ABCDE EABDC CDAEB DCBAE ACDEB BACED DECBA DECAB CEDAB DBECA CEDBA DAEBC DCAEB AEBDC ADCBE CDEAB EDBCA AEBCD BECDA CBEDA BACDE CEBDA BADCE BDCEA DCABE BDAEC BEACD BCEDA DBCAE DABEC BCDAE BCEAD CEABD BADEC EBACD CEADB ACEBD CADBE BECAD EADBC ADBEC ACBED EBCAD DABCE EACDB EABCD ABECD AECBD ECDBA BDEAC DAECB DBAEC DBCEA ABDEC AECDB CAEDB DACBE EADCB BDECA CABDE CBADE EDCBA BEDAC EDBAC EBCDA ADCEB ECABD ACDBE ABEDC BDACE BDCAE ECBDA BCDEA EBDAC CBDEA ECADB ECBAD CADEB DEABC BAEDC CEBAD BEADC EDABC CDBAE BCAED AEDBC DCEAB CDEBA ADECB DEBAC DCBEA AEDCB DBEAC CDABE BCADE CAEBD ACEDB CBAED EACBD ADEBC EBDCA CBEAD CABED ABDCE BAECD DBACE ABCED EDCAB CDBEA Number of permutations is 120