两两交换链表节点，为什么不能用head

djdgf4

这是我通过的代码，但是我之前18是

while(head != nullptr && (head->next) != nullptr)

然后报了错误：

Line 21: Char 35: runtime error: member access within null pointer of type 'ListNode' (solution.cpp)
SUMMARY: UndefinedBehaviorSanitizer: undefined-behavior prog_joined.cpp:30:35

这是为什么？

[C++] 纯文本查看 复制代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        ListNode* _HEAD = new ListNode(0);//虚拟头节点
        ListNode* fast = nullptr,*last = nullptr;
        _HEAD->next = head;
        last = _HEAD;
        while(last->next != nullptr && (last->next->next) != nullptr)
        {
            fast = last->next;
            ListNode* tmp = fast->next;
            last->next = tmp;
            fast->next = tmp->next;
            tmp->next = fast;

            last = last->next->next;

        }
        return _HEAD->next;
    }
};

while(head != nullptr && (head->next) != nullptr)

熊猫拍板砖

可以使用head

/* Definition for singly-linked list. type ListNode struct {     Val int     Next ListNode } /
func swapPairs(head ListNode) ListNode {  
   if head == nil || head.Next == nil{        return head    }  
dummy:=&ListNode{Next:head}  
prev:=dummy  
for head!=nil && head.Next!=nil{      
// 把第二个结点摘取出来      
next:=head.Next      
// 先让第一个结点Next指向第三个结点      
head.Next=next.Next      
// 然后第二个结点和第三个结点断开关系      
next.Next=head      
// prev 指向临时结点      
prev.Next=next      
// 结点交换        
prev = head      
head=head.Next    }  
return dummy.Next}

hdydy

[C++] 纯文本查看 复制代码

class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        ListNode* _HEAD = new ListNode(0, head);// 虚拟头节点
        ListNode* p = _HEAD, *q = nullptr;// p->head->q,交换head和q之后,p后移2步
        while(head && head->next) {
            q = head->next;
            head->next = q->next;
            q->next = head;
            p->next = q;
            p = head;
            head = p->next;
        }
        return _HEAD->next;
    }
};

laoyehuamao

太深奥了！！！