字典?wordList 中从单词 beginWord?和 endWord 的 转换序列 是一个按下述规格形成的序列?beginWord -> s1?-> s2?-> ... -> sk：
每一对相邻的单词只差一个字母。
?对于?1 <= i <= k?时，每个?si?都在?wordList?中。注意， beginWord?不需要在?wordList?中。
sk?== endWord
给你两个单词 beginWord?和 endWord 和一个字典 wordList ，返回 从?beginWord 到?endWord 的 最短转换序列 中的 单词数目 。如果不存在这样的转换序列，返回 0 。
?
示例 1：
输入：beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
输出：5
解释：一个最短转换序列是 "hit" -> "hot" -> "dot" -> "dog" -> "cog", 返回它的长度 5。
示例 2：
输入：beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
输出：0
解释：endWord "cog" 不在字典中，所以无法进行转换。
?
提示：
1 <= beginWord.length <= 10
endWord.length == beginWord.length
1 <= wordList.length <= 5000
wordList[i].length == beginWord.length
beginWord、endWord 和 wordList[i] 由小写英文字母组成
beginWord != endWord
wordList 中的所有字符串 互不相同
单向版本，时间152ms
class Solution {
public:
int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
unordered_set<string> wordlist(wordList.begin(), wordList.end());
if (wordlist.count(endWord) == 0)
return 0;
queue<string> q;
q.push(beginWord);
unordered_set<string> flag;
int step = 1;
while (!q.empty()) {
int size = q.size();
while (size--) {
string word = q.front();
q.pop();
for (int i = 0; i < word.size(); i++) {
for (int j = 'a'; j <= 'z'; j++) {
string temp = word;
if (j != word[i]) {
temp[i] = j;
if (wordlist.count(temp) == 0)
continue;
if (temp == endWord) {
return step + 1;
}
if (flag.count(temp) == 0) {
flag.insert(temp);
q.push(temp);
}
}
}
}
}
step++;
}
return 0;
}
};双向版本，时间28ms?
class Solution {
public:
int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
unordered_set<string> wordlist(wordList.begin(), wordList.end());
if (wordlist.count(endWord) == 0)
return 0;
queue<string> front;
queue<string> back;
front.push(beginWord);
back.push(endWord);
unordered_map<string, int> front_flag;
unordered_map<string, int> back_flag;
front_flag.insert(make_pair(beginWord, 1));
back_flag.insert(make_pair(endWord, 1));
while (!front.empty() && !back.empty()) {
int t = -1;
if (front.size() <= back.size()) {
t = update(front, front_flag, back_flag, wordlist);
}
else {
t = update(back, back_flag, front_flag, wordlist);
}
if (t != -1)
return t;
}
return 0;
}
int update(queue<string>& q, unordered_map<string, int>& flag, unordered_map<string, int>& other_flag, unordered_set<string>& wordlist) {
int size = q.size();
while (size--) {
string word = q.front();
q.pop();
for (int i = 0; i < word.size(); i++) {
for (int j = 'a'; j <= 'z'; j++) {
string temp = word;
if (j != word[i]) {
temp[i] = j;
if (wordlist.count(temp) == 0)
continue;
if (other_flag.count(temp) != 0) {
return flag[word] + other_flag[temp];
}
if (flag.count(temp) == 0) {
flag.insert(make_pair(temp, flag[word] + 1));
q.push(temp);
}
}
}
}
}
return -1;
}
};
