【算法刷题】leetcode word ladder_codewords 刷题-优快云博客

本文链接：https://blog.youkuaiyun.com/Neo_dot/article/details/80954237

本文介绍了一种利用广度优先搜索(BFS)算法求解两个单词间最短转换路径的方法。给定起始单词、目标单词及字典，通过逐层遍历字典中与当前层单词仅差一个字符的单词，实现从起始单词到目标单词的最短转换路径寻找。

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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 length5.

Note:

Return 0 if there is no such transformation sequence.
All words have the same length.
All words contain only lowercase alphabetic characters.

使用BFS搜索，即可以将dict分成若干层，每一层间的单词都与上一层相关的单词相差一个字符，从start开始匹配第一层，使用队列进行BFS，每次加入队列，就把单词从dict里删除，同时，要注意每一层的个数情况，进行while循环，并计数层数

    int ladderLength(string start, string end, unordered_set<string> &dict) {
		queue<string> strqueue;
		int res = 1;
		strqueue.push(start);
		while (!strqueue.empty())
		{
			int len = strqueue.size();
			while (len>0)
			{
				string frontstr = strqueue.front();
				strqueue.pop();
				len--;

				if (isDifOne(frontstr, end))
					return res+1;

				//BFS一轮，加了一层

			
				auto enditer = dict.end();
				for (auto iter = dict.begin(); iter != enditer;)
				{//此处要注意关联容器的删除，会影响迭代器，使迭代器失效
					if (isDifOne(frontstr, *iter))
					{

						strqueue.push(*iter);

						dict.erase(iter++);
					}
					else
					{
						++iter;
					}
				}
			}
            
			++res;
		}

		return 0;
        
    }
    bool isDifOne(string l, string r)
	{
		int sum(0);
		int length = l.length();
		for (int i = 0; i < length; ++i)
		{
			if (l[i] != r[i])
				++sum;
		}

		return sum == 1 ? true : false;
	}