import java.util.Iterator; import java.util.TreeSet; public class Sort { private String[] b = new String[] {"1","2","2","3","4","6"}; private int n = b.length; private boolean[] visited = new boolean[n]; private int[][] a = new int[n][n]; private String result = ""; private TreeSet<String> set = new TreeSet<String>(); public static void main(String[] args) { new Sort().start(); } private void start() { // Initial the map a[][] for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { if (i == j) { a[i][j] = 0; } else { a[i][j] = 1; } } } // 3 and 5 can not be the neighbor. a[3][5] = 0; a[5][3] = 0; // Begin to depth search. for (int i = 0; i < n; i++) { this.depthFirstSearch(i); } // Print result treeset. Iterator it = set.iterator(); while (it.hasNext()) { String string = (String) it.next(); System.out.println(string); } } private void depthFirstSearch(int startIndex) { visited[startIndex] = true; result = result + b[startIndex]; if (result.length() == n) { // "4" can not be the third position. if (result.indexOf(" 4 ") != 2) { // Filt the duplicate value. set.add(result); } } for (int j = 0; j < n; j++) { if (a[startIndex][j] == 1 && visited[j] == false) { depthFirstSearch(j); } } // restore the result value and visited value after listing a node. result = result.substring(0, result.length() - 1); visited[startIndex] = false; } }
俺也来贴一个 String input = "abc"; //求a/b/c三个字符的全排列 Set<String> results = new LinkedHashSet<String>(); results.add(""); for (int i=0; i<input.length(); i++) for (int j=0; j<input.length(); j++){ String current = input.substring(j, j+1); Set<String> newones = new LinkedHashSet<String>(); for(String s : results){ if (s.indexOf(current)==-1) s+=current; newones.add(s); } results.addAll(newones); } for (String s: results) if (s.length()==input.length())System.out.println(s);短一些,呵呵
String input = "abc"; //求a/b/c三个字符的全排列 Set<String> results = new LinkedHashSet<String>(); results.add(""); for (int i=0; i<input.length(); i++) for (int j=0; j<input.length(); j++){ String current = input.substring(j, j+1); Set<String> newones = new LinkedHashSet<String>(); for(String s : results){ if (s.indexOf(current)==-1) s+=current; newones.add(s); } results.addAll(newones); } for (String s: results) if (s.length()==input.length())System.out.println(s); 膜拜!
import java.util.Iterator;
import java.util.TreeSet; public class Sort { private String[] b = new String[] {"1","2","2","3","4","6"}; private int n = b.length; private boolean[] visited = new boolean[n]; private int[][] a = new int[n][n]; private String result = ""; private TreeSet<String> set = new TreeSet<String>(); public static void main(String[] args) {
new Sort().start();
} private void start() { // Initial the map a[][] for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == j) {
a[i][j] = 0;
} else {
a[i][j] = 1;
}
}
} // 3 and 5 can not be the neighbor. a[3][5] = 0;
a[5][3] = 0; // Begin to depth search. for (int i = 0; i < n; i++) {
this.depthFirstSearch(i);
} // Print result treeset. Iterator it = set.iterator();
while (it.hasNext()) {
String string = (String) it.next();
System.out.println(string);
}
} private void depthFirstSearch(int startIndex) {
visited[startIndex] = true;
result = result + b[startIndex];
if (result.length() == n) {
// "4" can not be the third position. if (result.indexOf(" 4 ") != 2) {
// Filt the duplicate value. set.add(result);
}
}
for (int j = 0; j < n; j++) {
if (a[startIndex][j] == 1 && visited[j] == false) {
depthFirstSearch(j);
}
} // restore the result value and visited value after listing a node. result = result.substring(0, result.length() - 1);
visited[startIndex] = false;
}
}
Set<String> results = new LinkedHashSet<String>();
results.add("");
for (int i=0; i<input.length(); i++)
for (int j=0; j<input.length(); j++){
String current = input.substring(j, j+1);
Set<String> newones = new LinkedHashSet<String>();
for(String s : results){
if (s.indexOf(current)==-1) s+=current;
newones.add(s);
}
results.addAll(newones);
}
for (String s: results)
if (s.length()==input.length())System.out.println(s);短一些,呵呵
String input = "abc"; //求a/b/c三个字符的全排列
Set<String> results = new LinkedHashSet<String>();
results.add("");
for (int i=0; i<input.length(); i++)
for (int j=0; j<input.length(); j++){
String current = input.substring(j, j+1);
Set<String> newones = new LinkedHashSet<String>();
for(String s : results){
if (s.indexOf(current)==-1) s+=current;
newones.add(s);
}
results.addAll(newones);
}
for (String s: results)
if (s.length()==input.length())System.out.println(s);
膜拜!