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Swap Nodes in Pairs

时间:2014-09-06 13:39:03      阅读:177      评论:0      收藏:0      [点我收藏+]

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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.

思路:

class Solution {
public:
    ListNode *swapPairs( ListNode *head ) {
        ListNode guard( -1 );
        guard.next = head;
        head = &guard;
        while( head ) {
            ListNode *node = head->next;
            if( !node || !node->next ) { return guard.next; }
            head->next = node->next;
            node->next = head->next->next;
            head->next->next = node;
            head = node;
        }
        return guard.next;
    }
};

 递归:

 1 class Solution {
 2 public:
 3     ListNode *swapPairs( ListNode *head ) {
 4         if( head == 0 || head->next == 0 ) { return head; }
 5         ListNode *node = head->next;
 6         head->next = swapPairs( node->next );
 7         node->next = head;
 8         return node;
 9     }
10 };

 

Swap Nodes in Pairs

标签:des   style   blog   color   io   ar   for   div   sp   

原文地址:http://www.cnblogs.com/moderate-fish/p/3959285.html

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