LeetCode 面试题06. 从尾到头打印链表

题目链接：https://leetcode-cn.com/problems/cong-wei-dao-tou-da-yin-lian-biao-lcof/

输入一个链表的头节点，从尾到头反过来返回每个节点的值（用数组返回）。

示例 1：

输入：head = [1,3,2]
输出：[2,3,1]
 

限制：

0 <= 链表长度 <= 10000

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     struct ListNode *next;
 6  * };
 7  */
 8 
 9 /**
10  * Note: The returned array must be malloced, assume caller calls free().
11  */
12 int* reversePrint(struct ListNode* head, int* returnSize){
13     if(head==NULL){
14         *returnSize=0;
15         return NULL;
16     }
17     int i,j,len=0;
18     struct ListNode *q=head;
19     while(q){
20         len++;
21         q=q->next;
22     }
23     *returnSize=len;
24     int *a=(int *)malloc(sizeof(int)*len);
25     q=head;
26     i=0;
27     while(q){
28         a[i++]=q->val;
29         q=q->next;
30     }
31     for(i=0;i<len/2;i++){
32         int tmp=a[i];
33         a[i]=a[len-i-1];
34         a[len-i-1]=tmp;
35     }
36     return a;
37 }

 另一种是链表先逆序：

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     struct ListNode *next;
 6  * };
 7  */
 8 
 9 /**
10  * Note: The returned array must be malloced, assume caller calls free().
11  */
12 int* reversePrint(struct ListNode* head, int* returnSize){
13     if(head==NULL){
14         *returnSize=0;
15         return NULL;
16     }
17     struct ListNode *pre=NULL,*cur=head,*tmp;
18     while(cur){
19         tmp=cur->next;
20         cur->next=pre;
21         pre=cur;
22         cur=tmp;
23     }
24     int i=0,len=0;
25     struct ListNode *q=pre;
26     while(q){
27         len++;
28         q=q->next;
29     }
30     *returnSize=len;
31     int *a=(int *)malloc(sizeof(int)*len);
32     q=pre;
33     while(q){
34         a[i++]=q->val;
35         q=q->next;
36     }
37     return a;
38 }

 

posted @ 2020-03-03 23:23  wydxry  阅读(344)  评论(0)  编辑  收藏  举报
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