Leetcode 24 Swap Nodes in Pairs

/**
* ID: 24
* Name: Swap Nodes in Pairs
* Data Structure: Linked List
* Time Complexity:
* Space Complexity:
* Tag: LinkList
* Difficult: Medium

* Problem:
* Given a linked list, swap every two adjacent nodes and return its head.
* 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.

思路一：

使用 pre, start , then方法，来处理反转，插入的操作，然后注意整个数组的长度是偶数还是奇数。

 1 class Solution {
 2 public:
 3     ListNode *swapPairs(ListNode *head) {
 4         if(head==NULL || head->next == NULL)
 5             return head; 
 6         ListNode * fakeNode = new ListNode(0);
 7         fakeNode -> next = head; 
 8         ListNode * pre = fakeNode;
 9         ListNode * start = pre->next; 
10         ListNode * then = start -> next; 
11         while(then)
12         {
13             start -> next = then -> next; 
14             then -> next = pre ->next; 
15             pre->next = then;
16             then = start -> next; 
17             pre = start; 
18             if(pre->next==NULL)
19                 break;
20             start = start ->next; 
21             then = start->next; 
22         }
23         return fakeNode->next; 
24     }
25 };

思路二：

 1 // recursive will be a little slower. 
 2 class Solution {
 3 public:
 4     ListNode *swapPairs(ListNode *head) {
 5         if(head==NULL || head->next == NULL)
 6             return head; 
 7         ListNode *next = head->next; 
 8         ListNode *nextnext = next->next;
 9         head ->next = this->swapPairs(nextnext);
10         next ->next = head; 
11         return next; 
12     }
13 };

posted on 2015-04-14 04:56 又岸阅读(140) 评论(0) 收藏举报