每日一题——两两交换链表中的节点

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* swapPairs(struct ListNode* head){
    //当链表为空或只有一个节点是，不要交换，直接返回原头节点
    if(head == NULL || head->next == NULL)
        return head;
    //建立新表头
    struct ListNode *newHead = (struct ListNode *)malloc(sizeof(struct ListNode));
    newHead->next = head;
    struct ListNode *cur = head, *prev = newHead;
    //当cur和cur->next都不为空时进行循环
    while(cur && cur->next)
    {
        struct ListNode *temp = cur->next;
        cur->next = temp->next;
        temp->next = cur;
        prev->next = temp;
        //交换完后更新prev和cur的位置
        prev = cur;
        cur = cur->next;
    }
    return newHead->next; //返回表头
}