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
    if (pHead1 == NULL && pHead2 == NULL) {
        return NULL;
    } else if (pHead1 == NULL || pHead2 == NULL) {
        return pHead1 ? pHead1 : pHead2;
    }
    struct ListNode* head = NULL;//变化的节点
    struct ListNode* retp = NULL;//头节点
  	//确定头节点
    if (pHead1->val <= pHead2->val) {
            head = pHead1;
            pHead1 = pHead1->next;
    } else {
        head = pHead2;
        pHead2 = pHead2->next;
    }
    retp = head;
    while(pHead1 && pHead2) {
        if (pHead1->val <= pHead2->val) {
            head->next = pHead1;
            pHead1 = pHead1->next;
        } else {
            head->next = pHead2;
            pHead2 = pHead2->next;
        }
        head = head->next;
    }
    head->next = pHead1 ? pHead1 : pHead2;//接上非空的链表
    return retp;
}