Word Ladder问题及解法

最新推荐文章于 2024-09-10 03:45:52 发布

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本文介绍了一种使用图遍历（BFS）算法解决单词转换问题的方法，即从一个初始单词通过改变一个字母逐步转换到目标单词的过程。文章详细阐述了如何构建图模型，设定visited标志数组记录已访问节点，并通过队列实现广度优先搜索，最终找到最短转换路径。

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问题描述：

Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that:

Only one letter can be changed at a time.
Each transformed word must exist in the word list. Note that beginWord is not a transformed word.

示例：

Given:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log","cog"]

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.
You may assume no duplicates in the word list.
You may assume beginWord and endWord are non-empty and are not the same.

问题分析：

对以此类问题，我们可以转换为一个图遍历的问题（BFS），图的每个节点间只有同一个位置上的字母不相同。设置visited标志数组，记录访问过的节点。wordList中包含了所有可以利用的节点。

过程详见代码：

class Solution {
public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
		unordered_set<string> visited;
		unordered_set<string> wordSet(wordList.begin(), wordList.end());
		int level = 1;
		int lastNum = 1;
		int curNum = 0;
		queue<string> que;
		que.push(beginWord);
		visited.insert(beginWord);
		while (!que.empty())
		{
			string cur = que.front();
			que.pop();
			lastNum--;
			for (int i = 0; i < cur.length(); i++)
			{
				for (char c = 'a'; c <= 'z'; c++)
				{
					if (c != cur[i])
					{
						string tcur = cur;
						tcur[i] = c;
						if (endWord == tcur && wordSet.find(tcur) != wordSet.end())
							return level + 1;
						if (wordSet.find(tcur) != wordSet.end() && visited.find(tcur) == visited.end())
						{
							curNum++;
							visited.insert(tcur);
							que.push(tcur);
						}
					}
				}
				
			}
            if (lastNum == 0)
				{
					lastNum = curNum;
					curNum = 0;
					level++;
				}
		}
		return 0;
	}

	
};