leetcode 676. Implement Magic Dictionary

原题：

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: NullInput: search("hello"), Output: FalseInput: search("hhllo"), Output: TrueInput: search("hell"), Output: FalseInput: search("leetcoded"), Output: False

Note:

1. You may assume that all the inputs are consist of lowercase letters a-z.
2. For contest purpose, the test data is rather small by now. You could think about highly efficient algorithm after the contest.
3. 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.

代码如下：

typedef struct {    char* string;    int len;    struct MagicDictionary* next;} MagicDictionary;/** Initialize your data structure here. */MagicDictionary* magicDictionaryCreate() {    MagicDictionary* head;    head=(MagicDictionary*)malloc(sizeof(MagicDictionary));    head->next=NULL;    return head;}/** Build a dictionary through a list of words */void magicDictionaryBuildDict(MagicDictionary* obj, char** dict) {    if(dict==NULL||*dict==NULL)        return;    //这个地方写的太脑残了    int dictLen=0;    char** q=dict;    while(*(q+dictLen)!=NULL)    {        dictLen++;    }    printf("%d",dictLen);    for(int n=0;n<dictLen;n++)    {        MagicDictionary* p;        p=(MagicDictionary*)malloc(sizeof(MagicDictionary));        int stringLen=strlen(*dict+n);        p->string=(char*)malloc(sizeof(char)*(stringLen+1));        strcpy(p->string,*(dict+n));        p->len=stringLen;        printf("%s",p->string);        p->next=obj->next;        obj->next=p;    }}/** Returns if there is any word in the trie that equals to the given word after modifying exactly one character */bool magicDictionarySearch(MagicDictionary* obj, char* word) {    int wordLen=strlen(word);    MagicDictionary* p=obj->next;    while(p!=NULL)    {        if(p->len!=wordLen)        {            p=p->next;        }        else        {            int diff=0;            for(int n=0;n<p->len;n++)            {                if(*(word+n)!=*(p->string+n))                    diff++;            }            if(diff==1)                return true;            else            {                p=p->next;            }        }    }    return false;}void magicDictionaryFree(MagicDictionary* obj) {    MagicDictionary* p=obj->next;    free(obj);    while(p!=NULL)    {        free(p->string);        MagicDictionary* temp=p;        p=temp->next;        free(temp);    }}/** * Your MagicDictionary struct will be instantiated and called as such: * struct MagicDictionary* obj = magicDictionaryCreate(); * magicDictionaryBuildDict(obj, dict); * bool param_2 = magicDictionarySearch(obj, word); * magicDictionaryFree(obj); */

我只能说leetcode对c太不友好了，欺负我们数据结构没有对应的属性或者函数是吧。。。

你tm让我怎么找二维数组的一维长度。。。。

难受。。。 搜了半天也没有好办法。