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| 1 | +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 |
| 2 | +那麼就可以用 BFS 找到最短路徑 |
| 3 | + |
| 4 | +// assume there is n strings in wordList , and each string length = m |
| 5 | +Time Complexity : O(mn^2) for O(n) nodes need O(nm) time to check if it exist in wordList and validity |
| 6 | +Space Complexity : O(n) for the queue |
| 7 | + |
| 8 | +class Solution { |
| 9 | +public: |
| 10 | + bool valid( string &s1 , string& s2 ) { |
| 11 | + int count = 0 ; |
| 12 | + for(int i = 0 ; i < s1.length() ; i++) |
| 13 | + if ( s1[i] != s2[i] ) |
| 14 | + count++ ; |
| 15 | + return count <= 1 ; |
| 16 | + } |
| 17 | + int ladderLength(string beginWord, string endWord, vector<string>& wordList) { |
| 18 | + int target = -1 ; |
| 19 | + for(int i = 0 ; i < wordList.size() ; i++) { |
| 20 | + if ( endWord == wordList[i] ) { |
| 21 | + target = i ; |
| 22 | + break ; |
| 23 | + } |
| 24 | + } |
| 25 | + |
| 26 | + queue<pair<int , int>> candidate ; |
| 27 | + wordList.push_back(beginWord) ; |
| 28 | + vector<int> finished(wordList.size() - 1) ; |
| 29 | + candidate.push( { wordList.size() - 1 , 1} ) ; |
| 30 | + |
| 31 | + while ( !candidate.empty() ) { |
| 32 | + string now = wordList[candidate.front().first] ; |
| 33 | + int distance = candidate.front().second ; |
| 34 | + candidate.pop() ; |
| 35 | + for( int i = 0 ; i < wordList.size() - 1 ; i++ ) { |
| 36 | + if ( finished[i] == 1 ) |
| 37 | + continue ; |
| 38 | + |
| 39 | + if ( valid(wordList[i] , now) ) { |
| 40 | + if ( i == target ) |
| 41 | + return distance + 1 ; |
| 42 | + else { |
| 43 | + candidate.push( {i , distance + 1} ) ; |
| 44 | + finished[i] = 1 ; |
| 45 | + } |
| 46 | + } |
| 47 | + } |
| 48 | + } |
| 49 | + return 0 ; |
| 50 | + } |
| 51 | +}; |
| 52 | + |
| 53 | +// another implementation |
| 54 | + |
| 55 | +class Solution { |
| 56 | +public: |
| 57 | + bool valid(string &s1, string &s2) { |
| 58 | + int len = s1.length() ; |
| 59 | + int count = 0 ; |
| 60 | + for(int i = 0 ; i < len ; i++) { |
| 61 | + if ( s1[i] != s2[i] ) |
| 62 | + count++ ; |
| 63 | + } |
| 64 | + return count == 1 ; |
| 65 | + } |
| 66 | + int ladderLength(string beginWord, string endWord, vector<string>& wordList) { |
| 67 | + queue<int> candidate ; |
| 68 | + |
| 69 | + int target = -1 ; |
| 70 | + vector<int> visited( wordList.size() ) ; |
| 71 | + for( int i = 0 ; i < wordList.size() ; i++) { |
| 72 | + if ( valid(beginWord , wordList[i]) ) { |
| 73 | + candidate.push(i) ; |
| 74 | + visited[i] = 1 ; |
| 75 | + } |
| 76 | + if ( endWord == wordList[i] ) |
| 77 | + target = i ; |
| 78 | + } |
| 79 | + |
| 80 | + if (target == -1) |
| 81 | + return 0 ; |
| 82 | + |
| 83 | + int step = 2 ; |
| 84 | + while ( !candidate.empty() ) { |
| 85 | + int size = candidate.size() ; |
| 86 | + while ( size-- ) { |
| 87 | + int index = candidate.front() ; |
| 88 | + candidate.pop() ; |
| 89 | + if (index == target) |
| 90 | + return step ; |
| 91 | + for(int i = 0 ; i < wordList.size() ; i++) { |
| 92 | + if ( valid(wordList[index] , wordList[i]) && visited[i] != 1) { |
| 93 | + candidate.push(i) ; |
| 94 | + visited[i] = 1 ; |
| 95 | + } |
| 96 | + } |
| 97 | + } |
| 98 | + step++ ; |
| 99 | + } |
| 100 | + |
| 101 | + return 0 ; |
| 102 | + } |
| 103 | +}; |
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