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.
改变指针:
if (!head || !(head->next) ) return head; ListNode *cur = head,*next = head->next; head = head->next; while(next){ next = next->next;//next->3 cur->next->next =cur;//2->1 if(!next || !(next->next)) { cur->next = next;//1->3; return head; } cur->next = next->next;//否则 1->4 cur = next;// cur ->3 next = next->next;//next->4 } return head;
只改变值:
ListNode* swapPairs(ListNode* head) { if(!head) return nullptr; ListNode *frontPtr=head,*backPtr=head->next; while(frontPtr && backPtr){ swap(frontPtr->val,backPtr->val); frontPtr=frontPtr->next; if(frontPtr!=nullptr) frontPtr=frontPtr->next; backPtr=backPtr->next; if(backPtr!=nullptr) backPtr=backPtr->next; } return head; }
递归:
class Solution { public: ListNode* swapPairs(ListNode* head) { if(head == NULL||head->next==NULL) return head; ListNode* next = head->next; head->next = swapPairs(next->next); next->next = head; return next; } };