循环链表基本操作

题目大致如下：

实现循环链表的插入删除操作

插入1-100个数，再按步进为2依次删除，3，5，7，9，13，15，17...直到链表为空，打印出依次删除的数

 

我当时没有完全做出来，现在已经实现，将代码贴到下面，不是最好的方法，欢迎大家讨论。

#include <stdio.h> #include <stdlib.h> #include <string.h> #include <windows.h> #define QUICK_DELETE typedef struct node { int data; struct node *next; }NodeT; void output(NodeT *Tail); NodeT *insert(NodeT **Tail, int data); NodeT *delete(NodeT **Tail, int data); //return the front ptr NodeT *quickdelete(NodeT *front, NodeT *current); NodeT *Tail = NULL; int main() { long begin, end; #ifdef QUICK_DELETE NodeT *front; #endif NodeT *head, *current; int i, step = 2; for(i = 1; i <= 100; i++) { insert(&Tail, i); } output(Tail); printf("---------------------------------------------------------------/n"); head = Tail->next; current = head; #ifdef QUICK_DELETE front = current; #endif begin = GetTickCount(); while(current != NULL) { while(step != 0) { #ifdef QUICK_DELETE front = current; #endif current = current->next; step--; } step = 2; //delete printf("%d ", current->data); #ifdef QUICK_DELETE current = quickdelete(front, current); #else current = delete(&Tail, current->data); #endif } end = GetTickCount(); printf("/n"); printf("end - begin = %d ms!/n", end - begin); system("pause"); return 0; } void output(NodeT *Tail) { NodeT *head = NULL, *current = NULL; if(Tail == NULL) return; head = Tail->next; current = head; while(current != Tail) { printf("%d ", current->data); current = current->next; } while(current != head->next) { printf("%d ", current->data); current = current->next; } printf("/n"); } NodeT *insert(NodeT **Tail, int data) { NodeT *newp = NULL; if(Tail == NULL) return NULL; newp = (NodeT *)malloc(sizeof(NodeT)); if(newp == NULL) return NULL; newp->data = data; newp->next = NULL; if(*Tail == NULL) { *Tail = newp; newp->next = newp; } else { newp->next = (*Tail)->next; (*Tail)->next = newp; } *Tail = (*Tail)->next; return (*Tail)->next; } NodeT *delete(NodeT **Tail, int data) { int find = 0; NodeT *head, *current, *front; if(Tail == NULL || *Tail == NULL) return NULL; head = (*Tail)->next; current = head; front = *Tail; while(current != *Tail) { if(current->data == data) { find = 1; break; } else { front = current; current = current->next; } } if((*Tail)->data == data) find = 1; if(find != 1) return NULL; if(head == *Tail) { free(*Tail); *Tail = front = current = NULL; } else { if(current == *Tail) { front->next = current->next; *Tail = front; free(current); } else { front->next = current->next; free(current); } } return front; } NodeT *quickdelete(NodeT *front, NodeT *current) { if(front == current && current == Tail) { free(current); return NULL; } else { if(current == Tail) { front->next = current->next; Tail = front; free(current); } else { front->next = current->next; free(current); } } return front; }