@@ -101,3 +101,66 @@ public:
101101 return 0 ;
102102 }
103103};
104+
105+ // 法二：因為此題 n > m，所以我們不使用找過所有字典的方法尋找合法字串，而是使用把所有可能的字串都嘗試過，出現在字典裡的我們就加入 queue 中
106+
107+ Time Complexity : O(mn) for O(n) nodes and each word will have O(26m) = O(m) valid words need to check
108+ Space Complexity : O(mn) for the hash map which store all the strings inside the wordList
109+ class Solution {
110+ public:
111+ bool valid(string &s1, string &s2) {
112+ int len = s1.length() ;
113+ int count = 0 ;
114+ for(int i = 0 ; i < len ; i++) {
115+ if ( s1[i] != s2[i] )
116+ count++ ;
117+ }
118+ return count == 1 ;
119+ }
120+ int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
121+ queue<int> candidate ;
122+
123+ int target = -1 ;
124+ unordered_map<string, int> appear ;
125+ for( int i = 0 ; i < wordList.size() ; i++) {
126+ appear[wordList[i]] = i ;
127+ if ( valid(beginWord , wordList[i]) ) {
128+ candidate.push(i) ;
129+ }
130+ if ( endWord == wordList[i] )
131+ target = i ;
132+ }
133+
134+ if (target == -1)
135+ return 0 ;
136+
137+ int step = 2 ;
138+ int len = beginWord.length() ;
139+ while ( !candidate.empty() ) {
140+ int size = candidate.size() ;
141+ while ( size-- ) {
142+ int index = candidate.front() ;
143+ candidate.pop() ;
144+ if (index == target)
145+ return step ;
146+
147+ for(int i = 0 ; i < len ; i++) {
148+ char old = wordList[index][i] ;
149+ for(char c = 'a' ; c <= 'z' ; c++ ) {
150+ if (c == old)
151+ continue ;
152+ wordList[index][i] = c ;
153+ if ( appear.find(wordList[index]) != appear.end() ) {
154+ candidate.push( appear[wordList[index]] ) ;
155+ appear.erase( wordList[index] ) ;
156+ }
157+ }
158+ wordList[index][i] = old ;
159+ }
160+ }
161+ step++ ;
162+ }
163+
164+ return 0 ;
165+ }
166+ };
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