求一個字符串的全部排列，網上找到的全部方法要麼是遞歸，要麼是三層for循環。 java
我這個方法不須要遞歸，只要兩層for循環，思路很是獨特。缺點是字符串長度較長時執行時間可能要幾萬年。但畢竟是本身的原創思路，且網上沒有這個算法，姑且記錄下來。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++) { // 字符串較長時須要改爲一個很大的數
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());
}
}
執行結果：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