LeetCode-127. Word Ladder

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:

1. Only one letter can be changed at a time.
2. Each transformed word must exist in the word list. Note thatbeginWordisnota transformed word.

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 assumebeginWordandendWordare non-empty and are not the same.

Example 1:

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

Output: 5

Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Example 2:

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

Output: 0

Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.

题解：

很经典的广度优先搜索。

一个pair队列保存当前状态和步长，按字母表搜索每个位置，找出每种wordList中有的可能，再放进队列中，逐一找到目标值。

class Solution {
public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        int res = INT_MAX, len = beginWord.length();
        queue<pair<string, int>> q;
        unordered_set<string> words(wordList.begin(), wordList.end());
        unordered_set<string> visit;
        q.push({beginWord, 1});
        visit.insert(beginWord);
        while (q.empty() == false) {
            auto cur = q.front();
            q.pop();
            if (cur.first == endWord) {
                res = min(res, cur.second);
                continue;
            }
            if (cur.second >= res) {
                continue;
            }
            for (int i = 0; i < len; i++) {
                string tmp = cur.first;
                for (int j = 0; j < 26; j++) {
                    tmp[i] = 'a' + j;
                    if (words.find(tmp) != words.end() && visit.find(tmp) == visit.end()) {
                        q.push({tmp, cur.second + 1});
                        visit.insert(tmp);
                    }
                }
            }
        }
        if (res == INT_MAX) {
            return 0;
        }
        return res;
    }
};