LeetCode126：Word Ladder

题目：

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.
• 解题思路：
• 这题一开始没啥思路，谷歌一下，才知要用到BFS，既然要用到BFS，那当然形成一个抽象图。这里我们将每一个字符串当做图中一节点，如果两字符串只需通过变化一个字符即可相等，我们认为这两字符串相连。
• 遍历图中节点时，我们通常会利用一个visit还标识是否访问过，这里我们将处理过的节点直接从dict中删除，以免重复处理。
• 实现代码：

#include <iostream>
#include <string>
#include <queue>
#include <unordered_set>
using namespace std;

/*
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.
*/
class Solution {
public:
    int ladderLength(string start, string end, unordered_set<string> &dict) {
        if(start.empty() || end.empty() || dict.empty())
            return 0;
        queue<string> squ[2];//这里需要用到两个队列，因为是bfs，按层遍历，所以需要一层一层进行处理
        squ[0].push(start);
        bool qid = false;
        int minLen = 1;
        while(!squ[qid].empty())
        {
            while(!squ[qid].empty())//处理同一层节点
            {
                string curstr = squ[qid].front();
                squ[qid].pop();
                for(int i = 0; i < curstr.size(); i++)
                {

                    for(char j = 'a'; j <= 'z'; j++)
                    {
                        if(j == curstr[i])
                            continue;
                        char t = curstr[i];
                        curstr[i] = j;
                        if(curstr == end)
                        {
                            return minLen+1;
                        }

                        if(dict.count(curstr) > 0)
                        {
                            squ[!qid].push(curstr);
                            dict.erase(curstr);
                        }
                        curstr[i] = t;
                    }

                }

            }
            qid = !qid;//表示将要处理的下一层
            minLen++;

        }
        return 0;

    }
};

int main(void)
{
    string start("hit");
    string end("cog");
    unordered_set<string> dict;
    dict.insert("hot");
    dict.insert("dot");
    dict.insert("dog");
    dict.insert("lot");
    dict.insert("log");
    Solution solution;
    int min = solution.ladderLength(start, end, dict);
    cout<<min<<endl;
    return 0;
}