1. 实现字符串按word逆序，假如字符串”This is Jack”，逆序为”Jack is This”。
解：第一步翻转整个字符串得到”kcaJ si sihT”，第二步翻转字符串中每个单词得到”Jack is This”。
void reverse(char *begin, char *end) { if (begin == NULL || end == NULL) { return; } while (begin < end) { // swap *begin and *end *begin = *begin + *end; *end = *begin - *end; *begin = *begin - *end; begin++; end--; } } char *reverse_words(char *data) { // check input if (data == NULL) { return NULL; } char *begin = data; char *end = data; // find the end of the data while (*end != ' ') { end++; } end = end - 1; // reverse data reverse(begin, end); // reverse word in data begin = end = data; while (*begin != ' ') { if (*begin == ' ') { begin++; end++; } else if (*end == ' ' || *end == ' ') { reverse(begin, --end); begin = ++end; } else { end++; } } return data; }2. 如何判断一个单链表是有环的？
解：设定两个指针slow和fast，slow每次移动一位，fast每次移动2位，如果有环路必定相遇既slow=fast
int is_exist_loop(struct List *head) { if (head == NULL) { return 0; } struct List *slow = head; struct List *fast = head; while (fast != NULL && fast->next != NULL) { slow = slow->next; fast = fast->next->next; if (slow == fast) { return 1; } } return 0; }3. 如何逆序单链表
解：将列表的下一个指针分别指向前一个，这里需注意在指正修改之前应该先保存下一个指针。
struct List *reverse_list(struct List *head) { struct List *new_head = NULL; struct List *cur = head; struct List *pre = NULL; while (cur != NULL) { struct List *next = cur->next; if (next == NULL) { new_head = cur; } cur->next = pre; pre = cur; cur = next; } }
