127. 单词接龙

127. 单词接龙

• 每一对相邻的单词只差一个字母。
•  对于 1 <= i <= k 时，每个 si 都在 wordList 中。注意， beginWord 不需要在 wordList 中。
• sk == endWord

给你两个单词 beginWord 和 endWord 和一个字典 wordList ，返回 从 beginWord 到 endWord 的 最短转换序列 中的 单词数目 。如果不存在这样的转换序列，返回 0 。

示例 1：

输入：beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
输出：5
解释：一个最短转换序列是 "hit" -> "hot" -> "dot" -> "dog" -> "cog", 返回它的长度 5。

示例 2：

输入：beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
输出：0
解释：endWord "cog" 不在字典中，所以无法进行转换。

提示：

• 1 <= beginWord.length <= 10
• endWord.length == beginWord.length
• 1 <= wordList.length <= 5000
• wordList[i].length == beginWord.length
• beginWord、endWord 和 wordList[i] 由小写英文字母组成
• beginWord != endWord
• wordList 中的所有字符串 互不相同

 1 #include 
 2 #include 
 3 #include 
 4 #include 
 5 #include <string>
 6 using namespace std;
 7 
 8 class Solution {
 9 public:
10     int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
11         if (beginWord.size() != endWord.size()) {
12             return 0;
13         }
14         if (wordList.size() == 0) {
15             return 0;
16         }
17         unordered_set<string> wordSet(wordList.begin(), wordList.end()); // vector转换成set,便于bfs时比较后将相邻元素入栈
18         unordered_set<string> visited; // 已访问单词集
19         queue<string> q;
20         q.push(beginWord);
21         int ans = 1;
22         while (!q.empty()) {
23             unsigned int size = q.size();
24             for (unsigned int i = 0; i < size; i++) { // 逐层遍历
25                 string now = q.front();
26                 q.pop();
27                 for (unsigned int j = 0; j < now.size(); j++) { // 遍历字符位置
28                     string next = now;
29                     for (unsigned int k = 0; k < 26; k++) {
30                         next[j] = k + 'a'; // 替换某个位置字符
31                         if (wordSet.count(next) > 0 && visited.count(next) == 0) {
32                             if (next == endWord) {
33                                 return ans + 1;
34                             }
35                             q.push(next);
36                             visited.emplace(next);
37                         }
38                     }
39                 }
40             }
41             ans++;
42         }
43         return 0;
44     }
45 };
46 
47 int main()
48 {
49     vector<string> wordList = {"hot", "dot", "dog", "lot", "log", "cog"};
50     string beginWord = "hit";
51     string endWord = "cog";
52     Solution *test = new Solution();
53     cout << test->ladderLength(beginWord, endWord, wordList) << endl;
54     system("pause");
55     return 0;
56 }