# Description

Given two words (beginWord and endWord), and a dictionary, find the length of shortest transformation sequence from beginWord to endWord, 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.

The original problem is here.

The original code is here.

# My Solution

I solve this problem in C++, as below:

``````/*
*Author: shuaijiang
*Email: zhaoshuaijiang8@gmail.com
*/
#include<iostream>
#include<string.h>
#include<stdlib.h>
using namespace std;

class Solution {
public:
int ladderLength(string beginWord, string endWord, unordered_set<string>& wordDict) {
map<string, int> wordVisited;
queue<string> curr;
bool found = false;
int len = 0;
curr.push(beginWord);
while(!curr.empty() && !found){
len ++;
queue<string> next;
while(!curr.empty() && !found){
string str = curr.front();
curr.pop();
for(int i=0;i<str.size();i++){
for(char ch='a';ch<='z';ch++){
if(ch == str[i])
continue;
swap(ch, str[i]);
if(str == endWord){
found = true;
break;
}
if(wordDict.count(str)>0 && wordVisited.find(str) == wordVisited.end()){
wordVisited[str] = 1;
next.push(str);
}

swap(str[i], ch);
}
}
}
curr = next;
}
if(found)
return len+1;
else
return 0;
}
};
``````

# Note

To solve the problem, width-first search was used.