题目:
递归实现,如果word[i]是通配符’.’,则从当前Trie结点所有子结点开始,搜索word.substr(i+1, n-i-1),而当word[i]=='.'并且i==n-1时,此时是最后一个结点,最后一个结点为通配符,因此只需要判断当前结点的children中有没有is_end==true的
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 #include <string> #include <vector> using std::string;using std::vector;typedef struct TrieNode { vector<struct TrieNode *> children; int data; bool is_end; } TrieNode; class WordDictionary { private : TrieNode *root; public : WordDictionary () { root = new TrieNode; root->children = vector <struct TrieNode *>(26 , nullptr ); root->is_end = false ; } void addWord (string word) { TrieNode *p = root, *node; int i = 0 , n = word.size (); int curr; for (i = 0 ; i < n; i++) { curr = word[i] - 'a' ; if (p->children[curr] != nullptr ) { p = p->children[curr]; continue ; } node = new TrieNode; node->children = vector <struct TrieNode *>(26 , nullptr ); node->data = curr; node->is_end = false ; p->children[curr] = node; p = node; } p->is_end = true ; } bool searchImpl (string word, TrieNode *root) { TrieNode *p = root; int i = 0 , n = word.size (), curr; for (i = 0 ; i < n; i++) { if (word[i] != '.' ) { curr = word[i] - 'a' ; if (p->children[curr] != nullptr ) { p = p->children[curr]; continue ; } return false ; } else { for (TrieNode *child : p->children) { if (child == nullptr ) { continue ; } if (i == n - 1 && child->is_end) { return true ; } if (searchImpl (word.substr (i + 1 , n - i - 1 ), child)) { return true ; } } return false ; } } return p->is_end; } bool search (string word) { return searchImpl (word, root); } };