字符串字典树判重Linux内核代码风格

来源：互联网发布：游戏辅助编程编辑：程序博客网时间：2024/05/29 19:02

背景：LeetCode - 127. Word Ladder

解决：

简单的广搜，用C++真的太好实现了，STL确实强大，但是用C语言来写呢？其实关键点就在于字符串如何判重，STL里的set可以搞，但是注意C里面没有这个东西，所以要么我们自己手写一个红黑树来实现set(红黑树不好写！很麻烦)，要么就使用字典树(这个确实相对来说好些得多)。写这个题其实也是自己想多学学C语言中一些高级的东西，所以就用Linux内核的代码风格来实现了下，不得不说相比较自己以前写的代码确实华丽多了,还有一点，我觉得很必要，就是内存的回收，虽然这个题我写了内存回收时间还长了，但是我个人觉得这是必要的，养成好习惯嘛。

代码：

typedef struct tagNode Node;typedef Node * pNode;typedef struct tagQueue Queue;typedef Queue *pQueue;typedef struct tagUsed Used;typedef Used *pUsed;typedef struct tagDicTree DicTree;typedef DicTree *pDicTree;typedef char *String;struct tagNode {    String string;    int step;};struct tagQueue {    pNode val;    pQueue next;};struct tagDicTree {    pDicTree alp[26];    int exsit;};struct tagUsed {    pDicTree root;    int (*search)(pDicTree root, const char *str, int len);    void (*insert)(pDicTree root, const char *str, int len);    void (*free_space)(pDicTree root);};#define INIT_STRUCT_USED(pused) (pused) = (pUsed)malloc(sizeof(Used));                          \                                (pused)->root = (pDicTree)malloc(sizeof(DicTree));              \                                memset((pused)->root, 0, sizeof(DicTree));                      \                                (pused)->search = search;                                       \                                (pused)->insert = insert;                                       \                                (pused)->free_space = free_space;                               int search(pDicTree root, const char *str, int len){    int i = 0;    pDicTree p = root;    for (; i < len; i++) {        int index = str[i] - 'a';        if (p->alp[index] != NULL)            p = p->alp[index];        else            break;    }    return i!= len ? 0 : (p->exsit == 1);}void insert(pDicTree root, const char *str, int len){    int i = 0;    pDicTree p = root;    for (; i < len; i++) {        int index = str[i] - 'a';        if (p->alp[index] == NULL) {            p->alp[index] = (pDicTree)malloc(sizeof(DicTree));            memset(p->alp[index], 0, sizeof(DicTree));            p->exsit = 0;        }        p = p->alp[index];    }    p->exsit = 1;}void free_space(pDicTree root){    int i;    if (root == NULL)        return ;    for (i = 0; i < 26; i++)        free_space(root->alp[i]);    free(root);}int check_bit_one(const char *str1, const char *str2, int len){    int ret = 0, i;    for (i = 0; i < len; i++)        if (str1[i] != str2[i])            ++ret;    return ret == 1;}int ladderLength(char* beginWord, char* endWord, char** wordList, int wordListSize) {    int len = strlen(beginWord);    int i;    pUsed used = NULL;    pQueue head = (pQueue)malloc(sizeof(Queue)), tail = head;    head->val = (pNode)malloc(sizeof(Node));    head->val->string = beginWord;    head->val->step = 1;    head->next = NULL;    INIT_STRUCT_USED(used);    while (head != NULL) {        pNode pre = head->val;        pQueue tQ = head;        if (!strcmp(pre->string, endWord))            return pre->step;        for (i = 0; i < wordListSize; i++) {            if (check_bit_one(pre->string, wordList[i], len) && !used->search(used->root, wordList[i], len)) {                tail->next = (pQueue)malloc(sizeof(Queue));                tail = tail->next;                tail->next = NULL;                tail->val = (pNode)malloc(sizeof(Node));                tail->val->string = wordList[i];                tail->val->step = pre->step + 1;                used->insert(used->root, wordList[i], len);            }        }        head = head->next;        free(tQ);    }    used->free_space(used->root);    return 0;}

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