Tuesday, August 9, 2016

Google – Password Sequence


Google – Password Sequence
有个保险箱,密码是四位0-9组成的数字,要求返回一个最短的暴力破解password sequence.
对于常见的保险箱,我们可以连续尝试保险箱的密码,比如当输入12345时,其实已经尝试了两个密码,1234和2345。而题目的意思就是找出一串最短的暴力破解序列,包含所有密码的可能性。
[Analysis]
从0000到9999一共有10000个数,也就是10000个密码。 由题意可以得到,最终最短的暴力破解序列长度应该为10003。
[Solution]
这道题其实是一道,Hamilton Cycle问题。每个vertex就是一个密码,如果当前密码的后三位数和另一个密码的前三位数一样,那么这两个节点之间就有一条edge. 每个vertex有10个adjacent nodes. 我们要做的就是在这个graph当中找出一条Hamilton Path.
但是实际的算法并没有那么复杂,并不需要另开数据结构来构建10000个节点的graph. 而是直接做backtracking.
而且仔细阅读这段code会发现其实也根本没有回溯,只是开了一个HashSet来保存遍历过的数字,不断尝试0000-9999之间的数字直到所有都遍历过为止。
  public String passwordSequence() {
    int start = 9999;
    String result = "" + start;
    Set<Integer> visited = new HashSet<>();
    visited.add(start);
     
    while (visited.size() < 10000) {
      int next = start * 10 % 10000;
      int i = 0;
      while (i <= 9) {
        if (visited.contains(next + i)) {
          i++;
        }
        else {
          break;
        }
      }
       
      visited.add(next + i);
      start = next + i;
      result += i;
    }
  }
同样地,对于 0 到 9 组成的全部四位数来说,我们可以设置 1000 个顶点,分别记作 000, 001, …, 999。如果某个数的后两位等于另一个数的前两位,就从前者出发,画一个箭头指向后者。容易想到,每个顶点的入度和出度都是 10,因此存在一条一笔画路径。也就是说,只需要按 10003 次数字键,便能试遍所有可能的四位数密码了。


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