Leetcode: Word Ladder

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start 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"]

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.

利用edit distance的思想，构造一颗树，根节点为start字符串，根的孩子节点为与其edit distance为1的字符串，以此类推，树中每一个父节点的孩子节点，都是与其edit distance为1的字符串。然后利用广度优先算法，逐层遍历该树，直到找到end为止。

#include 
#include 
#include 
#include 
using namespace std;

int ladderLength(string start,string end,unordered_set &dict)
{
	queue> q;
	q.push(make_pair(start,1));

	unordered_set visited;
	visited.insert(start);

	while (!q.empty())
	{
		string curStr = q.front().first;
		int curLevel = q.front().second;
		q.pop();

		for (int i=0;i dict;
	dict.insert("hot");
	dict.insert("dot");
	dict.insert("dot");
	dict.insert("lot");
	dict.insert("log");

	int len = ladderLength(start,end,dict);

	cout<