LeetCode24. Swap Nodes in Pairs(C++)
阿新 • • 發佈:2019-01-14
Given a linked list, swap every two adjacent nodes and return its head.
Example:
Given1->2->3->4
, you should return the list as2->1->4->3
.
Note:
- Your algorithm should use only constant extra space.
- You may not modify the values in the list's nodes, only nodes itself may be changed.
題目大意:兩兩交換節點
/** * 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* p,*q; p=head; if(p==NULL) return head; q=head->next; if(q==NULL) return head; while(q!=NULL&&q->next!=NULL){ int temp=p->val; p->val=q->val; q->val=temp; p=p->next->next; q=q->next->next; } if(p!=NULL&&q!=NULL){ int temp=p->val; p->val=q->val; q->val=temp; } return head; } };
另:直接交換節點的值也是可以的
/** * 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) { if(head==NULL) return head; ListNode *p,*q; p=head,q=head->next; while(q!=NULL){ swap(p->val,q->val); p=q->next; if(p==NULL) break; q=p->next; } return head; } };