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
 */
#include <stdlib.h>
struct ListNode* Merge(struct ListNode* pHead1, struct ListNode* pHead2 ) {
    // write code here
    #if 0
   if(pHead1 == NULL)
        return pHead2;
    else if ( pHead2 == NULL)
        return  pHead1;
    else if (pHead1 == NULL && pHead2 == NULL)
        return  NULL;
    struct ListNode *head = malloc(sizeof(struct ListNode));//新链表头结点
    struct ListNode *p = head;
    while (pHead1 != NULL && pHead2 != NULL)
    {
        if(pHead1 -> val >= pHead2 -> val) //pHead1链表节点值  大于 pHead2链表值  取 链表phead2节点
        {
            p->next = pHead2;
            pHead2 = pHead2->next;  //取完向后走一步
        }
        else //取phead1链表节点值
        {
            p->next = pHead1;
            pHead1 = pHead1->next;
        }
        p = p->next;
    }
    // if(pHead1)
    //     p->next = pHead1;
    // if(pHead2)
    //     p->next = pHead2;
    p->next = (pHead1?pHead1:pHead2);
    return head->next;//因为动态分配一个节点，并且next指向其他链表头结点，所以head值无效

    #endif
    //递归版本
    if(pHead1 == NULL)
        return pHead2;
    if(pHead2 ==NULL)
        return pHead1;
        // 根据值的大小决定哪个链表节点作为当前节点  
    if (pHead1->val <= pHead2->val) { 
        pHead1->next = Merge(pHead1->next, pHead2);  
        return pHead1;  
    } else {  
        pHead2->next = Merge(pHead1, pHead2->next);  //pHead2 设为当前节点 设置当前节点的下一个节点 递归
        return pHead2;  
    } 
}