word ladder

Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) fromstart to end, such that:

1. Only one letter can be changed at a time
2. Each intermediate word must exist in the dictionary

For example,

Given:
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.

思路参照路径查找，递归查找dict中和start相邻的下一个元素，但是不知道为什么在vs上运行的结果和网上提交的运行的不同

 

class Solution {
public:
bool isnext(string s1, string s2)//判断两个字符串是否是只有一个字母不同
{
int count = 0;

for (int i = 0, j = 0; i<s1.size(), j<s2.size(); i++, j++)
{
　　if (s1[i] != s2[j])
　　　　count++;
}
if (count == 1)
　　return true;
else
　　return false;
}
void findpath(string start, string end, unordered_set<string> &dict, vector<string>&path, 　　vector<vector<string>>&res)
　　{
　　path.push_back(start);
　　if (isnext(start, end))//到最后一个了
　　{
　　　　path.push_back(end);
　　　　res.push_back(path);
　　　　path.pop_back();
　　}

　　unordered_set<string>::iterator it;
　　if (dict.find(start) == dict.end())//这如果不讨论的话会导致递归进入死循环
　　　　it = dict.begin();
　　else
　　{
　　　　it = dict.find(start);

　　}

for (; it != dict.end(); it++)
{
　　if (isnext(start, *it))
{

     unordered_set<string> ndict (it, dict.end());
     findpath(*it, end, ndict, path, res);

     path.pop_back();
}
}
}
vector<vector<string>> findLadders(string start, string end, unordered_set<string> &dict) {
　　vector<string>path;
　　vector<vector<string>>res;
　　findpath(start, end, dict, path, res);
　　return res;
}
};