1 問題

反轉(zhuǎn)鏈表，比如0->1->2->3反轉(zhuǎn)后變成了3->2->1->0

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2 分析

搞3個指針，初始化一個指針，讓頭結(jié)點(diǎn)指向這里，然后另外一個指針初始化為NULL，然后讓第一個節(jié)點(diǎn)指向這里，然后頭結(jié)點(diǎn)依次向右移，這個初始化為NULL的指針也向右移動，然后最后當(dāng)頭結(jié)點(diǎn)的next指向NULL的時候，我們直接返回這個節(jié)點(diǎn)就行了。

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3 代碼實(shí)現(xiàn)

#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 nodeNode *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; }

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4 運(yùn)行結(jié)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

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