[LeedCode OJ]#24 Swap Nodes in Pairs

[LeedCode OJ]#24 Swap Nodes in Pairs

題意：

給你一個鏈表，要你對每兩個相鄰的節點進行交換

思路：

鏈表的結點不用想都知道通過next的指向來找，交換也是如此，通過改變指向就行了，但是注意要將尾結點的next指向空，否則要超時

/**
 * 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==nullptr || head->next==nullptr)
        return head;
        ListNode *newlist = new ListNode(0);
        ListNode *ptr = newlist;
        ListNode *cur = head;
        while(cur && cur->next)
        {
        	ListNode *pnext = cur->next->next;
        	ptr->next = cur->next;
        	ptr = ptr->next;
        	
        	ptr->next = cur;
        	ptr = ptr->next;
        	
        	ptr->next = nullptr;
        	cur  = pnext;
		}
		if(cur) ptr->next = cur;
		return newlist->next;
    }
};