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提到这个问题，我问：如何在 trie 中找到最长的单词？

我无法实现答案中给出的伪代码。

findLongest(trie):
 //first do a BFS and find the "last node"
 queue <- []
 queue.add(trie.root)
 last <- nil
 map <- empty map
while (not queue.empty()):
 curr <- queue.pop()
 for each son of curr:
    queue.add(son)
    map.put(son,curr) //marking curr as the parent of son
 last <- curr
//in here, last indicate the leaf of the longest word
//Now, go up the trie and find the actual path/string
curr <- last
str = ""
while (curr != nil):
      str = curr + str //we go from end to start   
    curr = map.get(curr)
return str

这就是我的方法

public static String longestWord (DTN d) {
  Queue<DTN> holding = new ArrayQueue<DTN>();
  holding.add(d);
  DTN last = null;
  Map<DTN,DTN> test = new ArrayMap<DTN,DTN>();
  DTN curr;
  while (!holding.isEmpty()) {
       curr = holding.remove();

      for (Map.Entry<String, DTN> e : curr.children.entries()) {
          holding.add(curr.children.get(e));
          test.put(curr.children.get(e), curr);
      }
          last = curr;

      }
  curr = last;
  String str = "";
  while (curr != null) {
      str = curr + str;
      curr = test.get(curr);

  }
  return str;

  }

我在以下位置收到 NullPointerException：

 for (Map.Entry<String, DTN> e : curr.children.entries())

如何找到并修复该方法的 NullPointerException 的原因，以便它返回 trie 中最长的单词？

4

2 回答 2

1

除了@Clark 的回答，请确保 curr.children 在取消引用之前不为空。

于 2012-11-04T04:57:47.123 回答
1

确保它curr.children.entries()不是空值。也许，DTN如果该节点没有子节点，则返回空值。这会导致您的NullPointerException.

在开始迭代之前尝试快速检查。

if(curr.children.entries() != null)
{
    //It's safe, so procede with going deeper into the trie.
}
于 2012-11-04T04:43:13.323 回答