【LeetCode】24. Swap Nodes in Pairs

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

Example:

Given 1->2->3->4, you should return the list as 2->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.

交换邻近的两个节点，但是不能用额外的链表空间，也不能修改节点的值来实现交换效果。

思路：加一个头结点，问题会好处理一些，剩下的就是指针的问题了。

/**
 * 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;
        if(head->next==NULL)
            return head;
        ListNode* left = head;
        ListNode* right=head->next;
        ListNode tem(0);
        ListNode* h = &tem;
        
        h->next = head;
        head = h;
            
        while(h->next!=NULL&&h->next->next!=NULL){
           
            left->next = right->next;
            right->next = left;
            h->next = right;
            h = left;
            if(h->next!=NULL&&h->next->next!=NULL){
                left = left->next;
                right = left->next;
            }
            
        }
        head = head->next;
        return head;
    }
};