leetCode 24. Swap Nodes in Pairs 鏈表

24. Swap Nodes in Pairs

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.

題目大意：

交換每兩個節(jié)點(diǎn)的位置。

代碼如下：

/**
 * 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* left,*right,*pre,*p;
        pre = NULL;//記錄每兩個節(jié)點(diǎn)前面的那個節(jié)點(diǎn)
        p = head;
        while(p !=NULL && p->next != NULL)
        {
            left = p;
            right = p->next;
            
            left->next = right->next;
            right->next = left;
            
            if(pre != NULL)
            {
                pre->next = right;
            }
            else//鏈表的頭兩個節(jié)點(diǎn)交換位置
            {
                head = right;
            }
            pre = left;
            p = left->next;
        }
        return head;
    }
};

2016-08-12 23:51:00