676. Implement Magic Dictionary

Implement a magic directory with buildDict, and search methods.

For the method buildDict, you’ll be given a list of non-repetitive words to build a dictionary.

For the method search, you’ll be given a word, and judge whether if you modify exactly one character into another character in this word, the modified word is in the dictionary you just built.

Example 1:

Input: buildDict(["hello", "leetcode"]), Output: Null
Input: search("hello"), Output: False
Input: search("hhllo"), Output: True
Input: search("hell"), Output: False
Input: search("leetcoded"), Output: False

Note:
- You may assume that all the inputs are consist of lowercase letters a-z.
- For contest purpose, the test data is rather small by now. You could think about highly efficient algorithm after the contest.
- Please remember to RESET your class variables declared in class MagicDictionary, as static/class variables are persisted across multiple test cases. Please see here for more details.

Approach

1. 题目大意实现类的方法，里面唯一难的地方应该就是search()方法实现，这里我是用深搜实现的，我的思路是用flag做标记，flag为true说明已经有字母用了移形换影，false说明还没用这技能，那么我每深搜一个字母的时候，我就判断flag，假如flag为true，那么已经用过这技能，那么我只能沿这个字母往下搜，假如flag为false，那么技能还没使用，我可以选择用，也可以选择不用，然后继续往下搜，递归边界就是当搜到最后一个字母的时候，也是根据flag做相应的判断，看代码思路会更加清晰。

Code

struct Trie
{
    Trie* letter[26];
    bool isword;
    Trie() {
        isword = false;
        for (int i = 0; i < 26; i++) {
            letter[i] = nullptr;
        }
    }
};
class MagicDictionary {
public:
    Trie* root;
    /** Initialize your data structure here. */
    MagicDictionary() {
        root = new Trie();
    }

    /** Build a dictionary through a list of words */
    void buildDict(vector<string> dict) {
        for (string &d : dict) {
            Trie *head = root;
            for (char &w : d) {
                int c = w - 'a';
                if (!head->letter[c]) {
                    head->letter[c] = new Trie();
                }
                head = head->letter[c];
            }
            head->isword = true;
        }
    }
    bool Search_str(string &word, int n, Trie *head, int pos, bool flag) {
        int c = word[pos] - 'a';
        if (pos == n) {
            if (flag) {
                return  head&&head->letter[c] && head->letter[c]->isword;
            }
            else {
                for (int i = 0; i < 26; i++) {
                    if (!head->letter[i] || c == i)continue;
                    if (head->letter[i]->isword)return true;
                }
                return false;
            }
        }
        if (flag) {
            return head&&head->letter[c] && Search_str(word, n, head->letter[c], pos + 1, flag);
        }
        else {
            for (int i = 0; i < 26; i++) {
                if (head->letter[i] || c == i)continue;
                if (Search_str(word, n, head->letter[i], pos + 1, true)) {
                    return true;
                }
            }
            return  head&&head->letter[c] && Search_str(word, n, head->letter[c], pos + 1, false);
        }
    }
    /** Returns if there is any word in the trie that equals to the given word after modifying exactly one character */
    bool search(string word) {
        return Search_str(word, word.size() - 1, root, 0, false);
    }
};