算法分析与设计第四周作业

算法分析与设计第四周作业

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题目

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.
题目传送门
题目大意为从beginWord开始，每次只能替换已经现在单词的其中一个字符，直到得到最终能否得到endWord。其中每次的变化不能是随意的，必须变为WordList中的其中一个单词才可以。

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解答

这道题其实就是画出图，用类似dijkstra的算法来解就很容易解决啦，代码如下。

class Solution {
public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        unordered_set<string> wordSet(wordList.begin(), wordList.end());
        unordered_map<string, int> pathCnt{{{beginWord, 1}}};
        queue<string> q{{beginWord}};
        while (!q.empty()) {
            string word = q.front(); q.pop();
            for (int i = 0; i < word.size(); ++i) {
                string newWord = word;
                for (char ch = 'a'; ch <= 'z'; ++ch) {
                    newWord[i] = ch;
                    if (wordSet.count(newWord) && newWord == endWord) return pathCnt[word] + 1;
                    if (wordSet.count(newWord) && !pathCnt.count(newWord)) {
                        q.push(newWord);
                        pathCnt[newWord] = pathCnt[word] + 1;
                    }   
                }
            }
        }
        return 0;
    }
};

解答详情传送门