标签:leetcode
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.
此题是LeetCode 25 Reverse Nodes in k-Group的一种特殊情况,即K=2. LeetCode25的解法请参考:http://blog.csdn.net/sunao2002002/article/details/46416977。故本题代码如下:
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: int getListSize(ListNode* head) { int count = 0; while (head) { head = head->next; count++; } return count; } ListNode* addHead(ListNode*head, ListNode*Node) { Node->next = head; return Node; } ListNode* reverseKGroup(ListNode* head, int k) { int length = getListSize(head); ListNode tmpHead(-1); ListNode *pNode = &tmpHead; while(length >= k){ ListNode* pHead = NULL; for (int i=0; i<k; i++) { ListNode*ptmpNode = head; head = head->next; ptmpNode->next = NULL; pHead = addHead(pHead, ptmpNode); } pNode->next = pHead; while(pNode->next) pNode = pNode->next; length -= k; } pNode->next = head; return tmpHead.next; } ListNode* swapPairs(ListNode* head) { return reverseKGroup(head, 2); } };
LeetCode 24:Swap Nodes in Pairs
标签:leetcode
原文地址:http://blog.csdn.net/sunao2002002/article/details/46486057