转载一个图数据结构题目

用1、2、2、3、4、5这六个数字，用java写一个main函数，打印出所有不同的排列，如：512234、412345等，要求："4"不能在第三位，"3"与"5"不能相连.

采用二维数组定义图结构，最后的代码是：

import java.util.Iterator;

import java.util.TreeSet;

public class TestQuestion {

private String[] b = new String[]{"1", "2", "2", "3", "4", "5"};

private int n = b.length;

private boolean[] visited = new boolean
;

private int[][] a = new int

;

private String result = "";

private TreeSet set = new TreeSet();

public static void main(String[] args) {

new TestQuestion().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();

// "4" can not be the third position.

if (string.indexOf("4") != 2) {

System.out.println(string);

}

}

}

private void depthFirstSearch(int startIndex) {

visited[startIndex] = true;

result = result + b[startIndex];

if (result.length() == n) {

// Filt the duplicate value.

set.add(result);

}

for(int j = 0; j < n; j++) {

if (a[startIndex][j] == 1 && visited[j] == false) {

depthFirstSearch(j);

} else {

continue;

}

}

// restore the result value and visited value after listing a node.

result = result.substring(0, result.length() -1);

visited[startIndex] = false;

}

}

转自 javaeye：http://www.javaeye.com/topic/55873