### 题目描述

A transformation sequence from word `beginWord` to word `endWord` using a dictionary `wordList` is a sequence of words `beginWord -> s1 -> s2 -> ... -> sk` such that:

• Every adjacent pair of words differs by a single letter.
• Every `si` for `1 <= i <= k` is in wordList. Note that `beginWord` does not need to be in wordList.
• `sk == endWord` Given two words, `beginWord` and `endWord`, and a dictionary `wordList`, return the number of words in the shortest transformation sequence from `beginWord` to `endWord`, or `0` if no such sequence exists.

### 例子

#### 例子 1

Input: `beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]` Output: `5` Explanation: `One shortest transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> cog", which is 5 words long.`

#### 例子 2

Input: `beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]` Output: `0` Explanation: `The endWord "cog" is not in wordList, therefore there is no valid transformation sequence.`

### Constraints

• `1 <= beginWord.length <= 10`
• `endWord.length == beginWord.length`
• `1 <= wordList.length <= 5000`
• `wordList[i].length == beginWord.length`
• `beginWord`, `endWord`, and `wordList[i]` consist of lowercase English letters.
• `beginWord != endWord`
• All the words in `wordList` are unique.

### 解题思路

``````#include <queue>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <vector>

class Solution {
public:
std::vector<std::string>& wordList) {
// construct graph
std::unordered_map<int, std::vector<int>> graph;
std::queue<int> candidates;
for (int i = 0; i < wordList.size(); ++i) {
if (wordList[i] == endWord) {
candidates.push(i);
}
graph[i] = {};
for (int j = 0; j < i; ++j) {
if (countDiff(wordList[i], wordList[j]) == 1) {
graph[i].push_back(j);
graph[j].push_back(i);
}
}
}

int step = 1;
std::unordered_set<int> used;
while (!candidates.empty()) {
std::queue<int> next;
while (!candidates.empty()) {
int idx = candidates.front();
candidates.pop();
// cout << wordList[idx] << endl;
if (countDiff(wordList[idx], beginWord) == 1) {
return ++step;
}
used.insert(idx);

for (auto neighbor : graph[idx]) {
if (used.count(neighbor) == 0) {
next.push(neighbor);
}
}
}
step++;
candidates = next;
}

return 0;
}

private:
int countDiff(const std::string& first, const std::string& second) {
if (first.size() != second.size()) {
return -1;
}

int count = 0;
for (int i = 0; i < first.size(); ++i) {
count += (first[i] != second[i]);
}

return count;
}
};
``````
• 时间复杂度: `O(|V| + |E|)`
• 空间复杂度: `O(|E|)`

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