Create Word Ladder

Author: AAAA-source

精選高頻HARD題目/Word Ladder

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+想法：可把每個字串都想像成是一節點，而字串間如果只有一個字母有差異的話就會有邊相連
+那麼就可以用 BFS 找到最短路徑
+
+// assume there is n strings in wordList , and each string length = m
+Time Complexity : O(mn^2) for O(n) nodes need O(nm) time to check if it exist in wordList and validity
+Space Complexity : O(n) for the queue
+
+class Solution {
+public:
+    bool valid( string &s1 , string& s2 ) {
+        int count = 0 ;
+        for(int i = 0 ; i < s1.length() ; i++) 
+            if ( s1[i] != s2[i] )
+                count++ ;
+        return count <= 1 ;
+    }
+    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
+        int target = -1 ;
+        for(int i = 0 ; i < wordList.size() ; i++) {
+            if ( endWord == wordList[i] ) {
+                target = i ;
+                break ;
+            }
+        }
+
+        queue<pair<int , int>> candidate ;
+        wordList.push_back(beginWord) ;
+       vector<int> finished(wordList.size() - 1) ;
+        candidate.push( { wordList.size() - 1 , 1} ) ;
+        
+        while ( !candidate.empty() ) {
+            string now = wordList[candidate.front().first] ;
+            int distance = candidate.front().second ;
+            candidate.pop() ;
+            for( int i = 0 ; i < wordList.size() - 1 ; i++ ) {
+                if ( finished[i] == 1 )
+                    continue ;
+                
+                if ( valid(wordList[i] , now) ) {
+                    if ( i == target )
+                        return distance + 1 ;
+                    else {
+                        candidate.push( {i , distance + 1} ) ;
+                        finished[i] = 1 ;
+                    }
+                }
+            }
+        }
+        return 0 ;
+    }
+};
+
+// another implementation
+
+class Solution {
+public:
+    bool valid(string &s1, string &s2) {
+        int len = s1.length() ;
+        int count = 0 ;
+        for(int i = 0 ; i < len ; i++) {
+            if ( s1[i] != s2[i] )
+                count++ ;
+        }
+        return count == 1 ;
+    }
+    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
+        queue<int> candidate ;
+        
+        int target = -1 ;
+        vector<int> visited( wordList.size() ) ;
+        for( int i = 0 ; i < wordList.size() ; i++) {
+            if ( valid(beginWord , wordList[i]) ) {
+                candidate.push(i) ;
+                visited[i] = 1 ;
+            }
+            if ( endWord == wordList[i] )
+                target = i ;
+        }
+
+        if (target == -1)
+            return 0 ;
+
+        int step = 2 ;
+        while ( !candidate.empty() ) {
+            int size = candidate.size() ;
+            while ( size-- ) {
+                int index = candidate.front() ;
+                candidate.pop() ;
+                if (index == target)
+                    return step ;
+                for(int i = 0 ; i < wordList.size() ; i++) {
+                    if ( valid(wordList[index] , wordList[i]) && visited[i] != 1) {
+                        candidate.push(i) ;
+                        visited[i] = 1 ;
+                    }
+                }
+            }
+            step++ ;
+        }
+
+        return 0 ;
+    }
+};