笔试算法题

笔试算法题

一著名软件公司的java笔试算法题！       
   算法程序题：       
           该公司笔试题就1个，要求在10分钟内作完。       
           题目如下：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。

循环方法
class Test2 {
public static void main(String args[]) {
   char[] c = { '1', '2', '2', '3', '4', '5' };
   for (int i = 0; i < c.length; i++) {
    for (int j = 0; j < c.length; j++) {
     if (i == j || (c[i] == '3' && c[j] == '5'))
      continue;
     for (int k = 0; k < c.length; k++) {
      if (k == i || k == j || (c[j] == '3' && c[k] == '5')
        || c[k] == '4') {
       continue;
      }
      for (int m = 0; m < c.length; m++) {
       if (m == k || m == i || m == j
         || (c[k] == '3' && c[m] == 5)) {
        continue;
       }
       for (int n = 0; n < c.length; n++) {
        if (n == m || n == k || n == i || n == j
          || (c[m] == '3' && c[n] == '5')) {
         continue;
        }
        for (int l = 0; l < c.length; l++) {
         if (l == n || l == m || l == k || l == j
           || l == i
           || (c[n] == '3' && c[l] == '5')) {
          continue;
         }
         System.out.println(c[i] + "" + c[j] + "" + c[k]
           + "" + c[m] + "" + c[n] + "" + c[l]);
        }
       }
      }
     }
    }
   }
}
}

 

递归方法
class Test2 {

public int[] getNextDifPos(int[] before, int add) {
   int[] res = new int[before.length + 1];
   for (int i = 0; i < before.length; i++) {
    res[i] = before[i];
   }
   res[res.length - 1] = add;
   return res;
}

public int getNextPos(int[] pos) {
   for (int i = 0; i < pos.length; i++) {
    if (pos[i] != -1) {
     return i;
    }
   }
   return -1;
}

public void getStr(int[] before, char[] chars) {
   if (before.length == chars.length) {
    StringBuffer str = new StringBuffer();
    for (int i = 0; i < before.length; i++) {
     str.append(chars[before[i]]);
    }
    // 加判断条件
    if (str.charAt(2) == '4' || str.indexOf("35") >= 0) {
     return;
    }
    System.out.println(str.toString());
   } else if (before.length < chars.length) {
    int[] pos = new int[chars.length];
    for (int i = 0; i < before.length; i++) {
     pos[before[i]] = -1;
    }
    int nextPos;
    while ((nextPos = getNextPos(pos)) != -1) {
     pos[nextPos] = -1;
     getStr(getNextDifPos(before, nextPos), chars);
    }
   }
}

public static void main(String args[]) {
   char[] c = { '1', '2', '2', '3', '4', '5' };
   int[] l = {};
   Test2 t = new Test2();
   t.getStr(l, c);
}
}

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