LeetCode Word Ladder

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:

/*
*Word Ladder
*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.