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我最近发布了一个关于优化算法以计算 Levenshtein 距离的问题，这些回复将我带到了关于Levenshtein 距离的 Wikipedia 文章。

文章提到，如果在最大距离上存在一个界限k，则可能的结果可以来自给定查询，那么运行时间可以从O(mn)减少到O(kn)，m和n是字符串。我查看了算法，但我无法真正弄清楚如何实现它。我希望在这里得到一些线索。

优化是“可能的改进”下的#4。

让我感到困惑的部分是说我们只需要计算以主对角线为中心的宽度为2k+1的对角线条纹（主对角线定义为坐标 (i,i)）。

如果有人可以提供一些帮助/见解，我将非常感激。如果需要，我可以在这里发布书中算法的完整描述作为答案。

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我已经做过很多次了。我这样做的方式是对可能变化的游戏树进行递归深度优先树遍历。有一个预算k的变化，我用它来修剪树。有了这个例程，首先我用 k=0，然后 k=1，然后 k=2 运行它，直到我受到打击或者我不想再高了。

char* a = /* string 1 */;
char* b = /* string 2 */;
int na = strlen(a);
int nb = strlen(b);
bool walk(int ia, int ib, int k){
  /* if the budget is exhausted, prune the search */
  if (k < 0) return false;
  /* if at end of both strings we have a match */
  if (ia == na && ib == nb) return true;
  /* if the first characters match, continue walking with no reduction in budget */
  if (ia < na && ib < nb && a[ia] == b[ib] && walk(ia+1, ib+1, k)) return true;
  /* if the first characters don't match, assume there is a 1-character replacement */
  if (ia < na && ib < nb && a[ia] != b[ib] && walk(ia+1, ib+1, k-1)) return true;
  /* try assuming there is an extra character in a */
  if (ia < na && walk(ia+1, ib, k-1)) return true;
  /* try assuming there is an extra character in b */
  if (ib < nb && walk(ia, ib+1, k-1)) return true;
  /* if none of those worked, I give up */
  return false;
}

添加解释特里搜索：

// definition of trie-node:
struct TNode {
  TNode* pa[128]; // for each possible character, pointer to subnode
};

// simple trie-walk of a node
// key is the input word, answer is the output word,
// i is the character position, and hdis is the hamming distance.
void walk(TNode* p, char key[], char answer[], int i, int hdis){
  // If this is the end of a word in the trie, it is marked as
  // having something non-null under the '\0' entry of the trie.
  if (p->pa[0] != null){
    if (key[i] == '\0') printf("answer = %s, hdis = %d\n", answer, hdis);
  }
  // for every actual subnode of the trie
  for(char c = 1; c < 128; c++){
    // if it is a real subnode
    if (p->pa[c] != null){
      // keep track of the answer word represented by the trie
      answer[i] = c; answer[i+1] = '\0';
      // and walk that subnode
      // If the answer disagrees with the key, increment the hamming distance
      walk(p->pa[c], key, answer, i+1, (answer[i]==key[i] ? hdis : hdis+1));
    }
  }
}
// Note: you have to edit this to handle short keys.
// Simplest is to just append a lot of '\0' bytes to the key.

现在，为了将其限制在预算范围内，如果 hdis 太大，就拒绝下降。

于 2010-07-07T15:47:54.367 回答