Leetcode_word-ladder(c++ version)

地址：http://oj.leetcode.com/problems/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.

思路：1. 广度优先遍历搜索到的一定是最短路径(广搜是dijstra的特例).
2. 一次广搜到结果后即可返回

设置visited防止回搜

参考代码：

class Solution {
public:
int ladderLength(string start, string end, unordered_set<string> &dict) {
if(dict.empty() || start == end)
return 0;
queue<string>strq;
queue<int>depq;
strq.push(start);
depq.push(1);
string cur, nxt;
int depth;
unordered_set<string> visited;
while(!strq.empty())
{
nxt = cur = strq.front();
strq.pop();
depth = depq.front();
depq.pop();
if(cur == end)
return depth;
for(int i = 0; i < cur.length(); ++i)
{
for(char ch = 'a'; ch<='z'; ++ch)
{
if(ch!=cur[i])
{
nxt[i] = ch;
if(dict.find(nxt)!=dict.end() && visited.find(nxt)==visited.end())
{
visited.insert(nxt);
strq.push(nxt);
depq.push(depth+1);
}
nxt = cur;
}
}
}
}
return 0;
}
};