标签:wordList word cog Ladder que beginWord endWord Word LeetCode
算法描述:
Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that:
- Only one letter can be changed at a time.
- Each transformed word must exist in the word list. Note that beginWord is not a transformed word.
Note:
- Return 0 if there is no such transformation sequence.
- All words have the same length.
- All words contain only lowercase alphabetic characters.
- You may assume no duplicates in the word list.
- You may assume beginWord and endWord are non-empty and are not the same.
Example 1:
Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"] Output: 5 Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog", return its length 5.
Example 2:
Input: beginWord = "hit" endWord = "cog" wordList = ["hot","dot","dog","lot","log"] Output: 0 Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.
解题思路:广度优先搜索。用一个队列存储可能的候选字符串。用set 构建字典,并去重。对于每一个字符串,逐一选替换每一位字符,如果发现字典中存在这样的字符,则将该字符串加入到队列中,并在字典中取出该字符串。
int ladderLength(string beginWord, string endWord, vector<string>& wordList) { unordered_set<string> dict(wordList.begin(),wordList.end()); if(dict.find(endWord)==dict.end()) return 0; queue<string> que; que.push(beginWord); int level = 0; while(!que.empty()){ level++; int levelSize = que.size(); for(int i=0; i < levelSize; i++){ string word = que.front(); que.pop(); for(int j =0; j < beginWord.size(); j++){ char orginal_char = word[j]; for(char c = 'a'; c <= 'z'; c++){ word[j] = c; if(word == endWord) return level+1; if(dict.find(word)==dict.end()) continue; que.push(word); dict.erase(word); } word[j]=orginal_char; } } } return 0; }
标签:wordList,word,cog,Ladder,que,beginWord,endWord,Word,LeetCode 来源: https://www.cnblogs.com/nobodywang/p/10352994.html
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