题解 | #NC33-合并两个排序的链表#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */

/**
 * 
 * @param pHead1 ListNode类 
 * @param pHead2 ListNode类 
 * @return ListNode类
 */
struct ListNode* Merge(struct ListNode* pHead1, struct ListNode* pHead2 ) {
    // write code here
    if(pHead1 == NULL)
        return pHead2;
    if(pHead2 == NULL)
        return pHead1;
    //创建虚拟头节点，虚拟头节点的next指向合并完成的链表
    struct ListNode *dummyHead = (struct ListNode *)malloc(sizeof(struct ListNode));
    //p节点指向排序过程中最后一个的节点，初始时指向虚拟头节点
    struct ListNode *p = dummyHead;
    
    while(pHead1 != NULL && pHead2 != NULL)
    {
        if(pHead1->val < pHead2->val)
        {
            p->next = pHead1;//step1时，dummyHead->1
            p = pHead1;//p指向1，依次类推
            pHead1 = pHead1->next;
        }
        else
        {
            p->next = pHead2;//step2时，dummyHead->1->2
            p = pHead2;//p指向2，依此类推
            pHead2 = pHead2->next;
        }
    }
    
    if(pHead1 != NULL)
        p->next = pHead1;
    if(pHead2 != NULL)
        p->next = pHead2;
    
    return dummyHead->next;
}