题解 | #合并两个排序的链表#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param pHead1 ListNode类 
 * @param pHead2 ListNode类 
 * @return ListNode类
 */
struct ListNode* Merge(struct ListNode* pHead1, struct ListNode* pHead2 ) {
    // write code here
    struct ListNode* traversal1 = pHead1; //遍历pHead1的指针
    struct ListNode* traversal2 = pHead2; //遍历pHead2的指针
    struct ListNode* last = NULL;
    if(pHead1 == NULL) //当链表1不存在时
    {
        return pHead2;
    }
    if(pHead2 == NULL) //当链表2不存在时
    {
        return pHead1;
    }
    if(traversal1->val<traversal2->val) //确定链表头结点，返回头结点
    {
        last = traversal1;
        traversal1 = traversal1->next;
        while(1)
        {
            if(traversal1 == NULL) //链表1遍历完成
            {
                last->next = traversal2;
                return pHead1;
            }
            if(traversal2 == NULL) //链表2遍历完成
            {
                last->next = traversal1;
                return pHead1;
            }
            if(traversal1->val<traversal2->val)
            {
                last->next = traversal1;
                last = traversal1;
                traversal1 = traversal1->next;
            }
            else
            {
                last->next = traversal2;
                last = traversal2;
                traversal2 = traversal2->next;
            }
            
        }
    }
    else
    {
        last = traversal2;
        traversal2 = traversal2->next;
        while(1)
        {
            if(traversal1 == NULL)
            {
                last->next = traversal2;
                return pHead2;
            }
            if(traversal2 == NULL)
            {
                last->next = traversal1;
                return pHead2;
            }
            if(traversal1->val<traversal2->val)
            {
                last->next = traversal1;
                last = traversal1;
                traversal1 = traversal1->next;
            }
            else
            {
                last->next = traversal2;
                last = traversal2;
                traversal2 = traversal2->next;
            }
            
        }
    }
    
}