23
04/2015
[LeetCode] Swap Nodes in Pairs
Swap Nodes in Pairs
Given a linked list, swap every two adjacent nodes and return its head.
For example,
Given 1->2->3->4, you should return the list as 2->1->4->3.
Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.
解题思路：
注意头结点的不同。建议先创造一个伪头结点，其next指向head，这样，之后所有的操作都是一样的了。这样能够简化代码。
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* swapPairs(ListNode* head) {
ListNode* myHead=new ListNode(0);
myHead->next=head;
ListNode* pre=myHead;
while(pre->next!=NULL&&pre->next->next!=NULL){
ListNode* p=pre->next;
pre->next=p->next;
p->next=pre->next->next;
pre->next->next=p;
pre=pre->next->next;
}
head=myHead->next;
delete myHead;
return head;
}
};二次刷题（2015-08-18）
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* swapPairs(ListNode* head) {
ListNode* myHead = new ListNode(0);
myHead->next = head;
ListNode* p = myHead, *q;
while(p->next != NULL && p->next->next!=NULL){
q = p->next->next;
p->next->next = q->next;
q->next=p->next;
p->next = q;
p=p->next->next;
}
p = myHead->next;
delete myHead;
return p;
}
};转载请注明：康瑞部落 » [LeetCode] Swap Nodes in Pairs
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