LeetCode 24 (两两交换链表中的节点)

LeetCode 24 Medium

解题思路：

• 设置头节点headPoint 统一处理各种边界问题
• 设置pre指针指向要交换的节点的前驱，设置tmp指针指向和当前节点（current指针）进行交换，并将当前节点的next记录到node指针中
• 两两交换，直到current或者current -->next != NULL跳出循环，返回headPoint的next
• 时间复杂度：O(n)
• 空间复杂度：O(1)

结果：

代码

/**
 * 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 || head->next == NULL){
            return head;
        }
        ListNode* headPoint = (ListNode *)malloc(sizeof(ListNode));
        headPoint->next = head;
        ListNode* current = headPoint -> next;
        ListNode* tmp;
        ListNode* node;
        ListNode* swapNode;
        ListNode* pre=headPoint;
        while(current != NULL && current->next != NULL){
            tmp = current->next;
            node = tmp->next;
            pre->next = tmp;
            tmp->next = current;
            current->next = node;
            pre = current;
            current = current->next;
        }
        return headPoint->next;
    }
};