Word Ladder系列

Posted 2020-08-15 SarahLiu77

 

1.Word Ladder

问题描述：

给两个word（beginWord和endWord）和一个字典word list，找出从beginWord到endWord之间的长度最长的一个序列，条件：

1.字典中的每个单词只能使用一次；

2.序列中的每个单词都必须是字典中的单词；

例如：

Given:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

注意：

　　如果找不到合适的序列，返回0；

　　所有的单词长度都是一样的；

     所有的单词都只由小写字母组成。

思路：

采用DFS依次遍历

代码如下：

 

 1 class Solution {
 2 public:
 3     int ladderLength(string beginWord, string endWord, unordered_set<string>& wordDict) {
 4         unordered_set<string> s1 = {beginWord}; // Front end
 5         unordered_set<string> s2 = {endWord}; // Back end
 6         wordDict.erase(beginWord);
 7         wordDict.erase(endWord);
 8 
 9         return ladderLength(s1, s2, wordDict, 1);
10     }
11     
12 private:
13     int ladderLength(unordered_set<string>& s1, unordered_set<string>& s2, unordered_set<string>& wordDict, int level) {
14         if (s1.empty()) // We can‘t find one.
15             return 0;
16         unordered_set<string> s3; // s3 stores all words 1 step from s1.
17         for (auto word : s1) {
18             for (auto& ch : word) {
19                 auto originalCh = ch;
20                 for (ch = ‘a‘; ch <= ‘z‘; ++ ch) {
21                     if (ch != originalCh) {
22                         if (s2.count(word))  // We found one.
23                             return level + 1;
24                         if (wordDict.count(word)) {
25                             wordDict.erase(word); // Avoid duplicates.
26                             s3.insert(word);
27                         }
28                     }
29                 }
30                 
31                 ch = originalCh;
32             }
33         }
34         // Continue with the one with smaller size.    
35         return (s2.size() <= s3.size()) ? ladderLength(s2, s3, wordDict, level + 1) : ladderLength(s3, s2, wordDict, level + 1);
36     }
37 };