【LeetCode】127. Word Ladder

Word Ladder

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 =

"hit"

end =

"cog"

dict =

["hot","dot","dog","lot","log"]

As one shortest transformation is

"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.

最短搜索路径，所以是广度优先搜索(BFS)。

有几个问题需要注意：

1、怎样判断是否为邻居节点？

按照定义，存在一个字母差异的单词为邻居，因此采用逐位替换字母并查找字典的方法寻找邻居。

（逐个比对字典里单词也可以做，但是在这题的情况下，时间复杂度会变高，容易TLE）

2、怎样记录路径长度？

对队列中的每个单词记录路径长度。从start进入队列记作1.

长度为i的字母的邻居，如果没有访问过，则路径长度为i+1

struct Node
{
string word;
int len;

Node(string w, int l): word(w), len(l) {}
};

class Solution {
public:
int ladderLength(string beginWord, string endWord, unordered_set<string>& wordDict) {
queue<Node*> q;
unordered_map<string, bool> m;
Node* beginNode = new Node(beginWord, 1);
q.push(beginNode);
m[beginWord] = true;
while(!q.empty())
{
Node* frontNode = q.front();
q.pop();
string frontWord = frontNode->word;
//neighbor search
for(int i = 0; i < frontWord.size(); i ++)
{
for(char c = 'a'; c <= 'z'; c ++)
{
if(c == frontWord[i])
continue;

string frontWordCp = frontWord;
frontWordCp[i] = c;
//end
if(frontWordCp == endWord)
return frontNode->len+1;
if(wordDict.find(frontWordCp) != wordDict.end() && m[frontWordCp] == false)
{
Node* neighborNode = new Node(frontWordCp, frontNode->len+1);
q.push(neighborNode);
m[frontWordCp] = true;
}
}
}
}
return 0;
}
};