Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:
- Only one letter can be changed at a time
- Each intermediate word must exist in the dictionary
For example,
Given:
start =
end =
dict =
start =
"hit"end =
"cog"dict =
["hot","dot","dog","lot","log"]
As one shortest transformation is
return its length
"hit" -> "hot" -> "dot" -> "dog" -> "cog",return its length
5.
Note:
- Return 0 if there is no such transformation sequence.
- All words have the same length.
- All words contain only lowercase alphabetic characters.
Word Ladder II
Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that:
- Only one letter can be changed at a time
- Each intermediate word must exist in the dictionary
For example,
Given:
start =
end =
dict =
start =
"hit"end =
"cog"dict =
["hot","dot","dog","lot","log"]
Return
[
["hit","hot","dot","dog","cog"],
["hit","hot","lot","log","cog"]
]
Note:
- All words have the same length.
- All words contain only lowercase alphabetic characters.
思路：
LeetCode中为数不多的考图的难题。尽管题目看上去像字符串匹配题，但从“shortest transformation sequence from start to end”还是能透露出一点图论中最短路径题的味道。如何转化？
1. 将每个单词看成图的一个节点。
2. 当单词s1改变一个字符可以变成存在于字典的单词s2时，则s1与s2之间有连接。
3. 给定s1和s2，问题I转化成了求在图中从s1->s2的最短路径长度。而问题II转化为了求所有s1->s2的最短路径。
无论是求最短路径长度还是求所有最短路径，都是用BFS。在BFS中有三个关键步骤需要实现:
1. 如何找到与当前节点相邻的所有节点。
这里可以有两个策略：
(1) 遍历整个字典，将其中每个单词与当前单词比较，判断是否只差一个字符。复杂度为：n*w，n为字典中的单词数量，w为单词长度。
(2) 遍历当前单词的每个字符x，将其改变成a~z中除x外的任意一个，形成一个新的单词，在字典中判断是否存在。复杂度为：26*w，w为单词长度。
这里可以和面试官讨论两种策略的取舍。对于通常的英语单词来说，长度大多小于100，而字典中的单词数则往往是成千上万，所以策略2相对较优。
2. 如何标记一个节点已经被访问过，以避免重复访问。
可以将访问过的单词从字典中删除。
3. 一旦BFS找到目标单词，如何backtracking找回路径？
Word Ladder I
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 | class Solution { public: int ladderLength(string start, string end, unordered_set<string> &dict) { dict.insert(end); queue<pair<string,int>> q; q.push(make_pair(start,1)); while(!q.empty()) { string s = q.front().first; int len = q.front().second; if(s==end) return len; q.pop(); vector<string> neighbors = findNeighbors(s, dict); for(int i=0; i<neighbors.size(); i++) q.push(make_pair(neighbors[i],len+1)); } return 0; } vector<string> findNeighbors(string s, unordered_set<string> &dict) { vector<string> ret; for(int i=0; i<s.size(); i++) { char c = s[i]; for(int j=0; j<26; j++) { if(c=='a'+j) continue; s[i] = 'a'+j; if(dict.count(s)) { ret.push_back(s); dict.erase(s); } } s[i] = c; } return ret; } }; |
