leetCode_Add to List 206. Reverse Linked List
2017-10-17 13:14
405 查看
总觉得犯得错误低级到不行
这次的题目很简单,给你一个链表的头指针(没有头节点),逆置这个链表,例如把 1->2->3->4 变成 4->3->2->1
拿到题目一看,这还不简单,先找到尾节点,不停的把头结点进行尾插法不就行了(头指针和尾指针相等时就停止),于是代码如下:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseList(ListNode* head) {
if (!head)
return head;
ListNode* rear = NULL,*temporary = NULL;
temporary = head;
while(temporary->next){//遍历直到为节点
temporary = temporary->next;
}
rear = temporary;//尾指针指向尾节点
while(head != rear){
temporary = head;
temporary->next = rear->next;
rear->next = temporary;
head = head->next;
}
return rear;
}
};看起来没什么问题,提交走起
结果:
Line 25: member access within null pointer of type 'struct ListNode'
而会出现这个错误是因为非法访问内存,试图读取NULL的成员,而错误又是在 temporary->next = rear->next; 那一定是temporary在某次被置为了NULL,一直在肉眼debug没找到,后来在本地debug了一遍,结果是在第一次while循环中最后一句 head = head->next 使 head 变为了NULL,第二次试图访问NULL->next这当然是不行的 ,问题找到了,那来看看我为什么会出现这个问题,原因就在 temporary = head;当把head赋给temporary后,它俩就指向了同一个节点,而后面我用temporay去修改节点的next域后,head->next也跟着改变了(开始写的时候竟然妄想他们是独立的)
下面给出修改后的代码
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseList(ListNode* head) {
if (!head)
return head;
ListNode* rear = NULL,*temporary = NULL;
temporary = head;
while(temporary->next){
temporary = temporary->next;
}
rear = temporary;
while(head != rear){
temporary = head->next;
head->next = rear->next;
rear->next = head;
head = temporary;
}
return rear;
}
};
有temporary去存head->next,在修改head最后把temporary赋给head就避开这种问题了。
这次的题目很简单,给你一个链表的头指针(没有头节点),逆置这个链表,例如把 1->2->3->4 变成 4->3->2->1
拿到题目一看,这还不简单,先找到尾节点,不停的把头结点进行尾插法不就行了(头指针和尾指针相等时就停止),于是代码如下:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseList(ListNode* head) {
if (!head)
return head;
ListNode* rear = NULL,*temporary = NULL;
temporary = head;
while(temporary->next){//遍历直到为节点
temporary = temporary->next;
}
rear = temporary;//尾指针指向尾节点
while(head != rear){
temporary = head;
temporary->next = rear->next;
rear->next = temporary;
head = head->next;
}
return rear;
}
};看起来没什么问题,提交走起
结果:
Line 25: member access within null pointer of type 'struct ListNode'
而会出现这个错误是因为非法访问内存,试图读取NULL的成员,而错误又是在 temporary->next = rear->next; 那一定是temporary在某次被置为了NULL,一直在肉眼debug没找到,后来在本地debug了一遍,结果是在第一次while循环中最后一句 head = head->next 使 head 变为了NULL,第二次试图访问NULL->next这当然是不行的 ,问题找到了,那来看看我为什么会出现这个问题,原因就在 temporary = head;当把head赋给temporary后,它俩就指向了同一个节点,而后面我用temporay去修改节点的next域后,head->next也跟着改变了(开始写的时候竟然妄想他们是独立的)
下面给出修改后的代码
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseList(ListNode* head) {
if (!head)
return head;
ListNode* rear = NULL,*temporary = NULL;
temporary = head;
while(temporary->next){
temporary = temporary->next;
}
rear = temporary;
while(head != rear){
temporary = head->next;
head->next = rear->next;
rear->next = head;
head = temporary;
}
return rear;
}
};
有temporary去存head->next,在修改head最后把temporary赋给head就避开这种问题了。
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