Google – Correct Misspelling


Google – Correct Misspelling
Given a dictionary and a misspelling word with only ONE misspelled character, correct it.
Ex: [apple, people, …], adple => apple
[Solution]
[Solution #1] Brute Force
Compare the misspelled word with every word in the dictionary.
time: O(n * Avg(len))
[Solution #2] One Edit Distance
Do one edit distance on every word in the dict with the misspelled word
其实就是Solution #1
time: O(n * Avg(len))
[Solution #3] Trie tree with k tolerant
这个没有具体实现,不过这种思想倒是没见过,其他题目或许可以借鉴。
思路就是对dict建trie tree,然后对misspelled word做dfs search, 不同于普通的trie tree搜索,这里可以设置一个容忍度,比如这题可以设为1,如果搜索过程中有1个character不同,可以继续往下搜索,而不是直接return。
不过这样的做法对于这道题而言时间复杂度并没有提升,依然是O(n * Avg(len))
[Solution #4] backtracking
因为每个character只有26种可能性,对Misspelled word的每个character尝试所有可能性看有没有一种可能性在dictionary里就可以了。
time: O(len * 26)
下面的code是backtracking的做法。
  public List<String> correctWord(String[] dict, String word) {
    List<String> result = new ArrayList<>();
    if (dict == null || dict.length == 0 || word == null) {
      return result;
    }
 
    Set<String> set = new HashSet<>();
    for (String s : dict) {
      set.add(s);
    }
 
    char[] ch = word.toCharArray();
    for (int i = 0; i < ch.length; i++) {
      char save = ch[i];
      for (char c = 'a'; c <= 'z'; c++) {
        if (c != ch[i]) {
          ch[i] = c;
          String newStr = new String(ch);
          if (set.contains(newStr)) {
            result.add(newStr);
          }
        }
      }
      ch[i] = save;
    }
 
    return result;
  }
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