1 问题
反转链表，比如0->1->2->3反转后变成了3->2->1->0
2 分析
搞3个指针，初始化一个指针，让头结点指向这里，然后另外一个指针初始化为NULL，然后让第一个节点指向这里，然后头结点依次向右移，这个初始化为NULL的指针也向右移动，然后最后当头结点的next指向NULL的时候，我们直接返回这个节点就行了。
3 代码实现
#include <stdio.h> typedef struct Node { int val; struct Node *next; } Node; /* *print list */ void print_list(Node *head) { if (head == NULL) { printf("head is NULL\n"); return; } Node *p = head; while (p != NULL) { printf("value is %d\n", p->val); p = p->next; } } /* *reverse list */ struct Node* reverse(Node *head) { if (head == NULL) { printf("reverse head is NULL\n"); return NULL; } Node *end = NULL; Node *p = head; Node *start = NULL; while (p != NULL) { //next node Node *next = p->next; //If next is NULL, we will store p, we will return p in the end; if (next == NULL) { end = p; } p->next = start; start = p; p = next; } return end; } int main() { //0->1->2->3; Node head, node1, node2, node3; head.val = 0; head.next = &node1; node1.val = 1; node1.next = &node2; node2.val = 2; node2.next = &node3; node3.val = 3; node3.next = NULL; print_list(&head); printf("list will reverse\n"); Node *hello = reverse(&head); print_list(hello); return 0; }
4 运行结果
value is 0 value is 1 value is 2 value is 3 list will reverse value is 3 value is 2 value is 1 value is 0
