题解 | #合并两个排序的链表#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 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
     */
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2) {
        // write code here
        ListNode* p;
        ListNode* h;
	  //特殊情况讨论
        if(!pHead1){
            h = pHead2;
            return h;
        }
        if(!pHead2){
            h = pHead1;
            return h;
        }
	  //初始情况，没有头节点的链表
        if(pHead1->val <= pHead2->val){
            p = pHead1;
            pHead1 = pHead1->next;
        }else{
            p = pHead2;
            pHead2 = pHead2->next;
        }
        h = p;
        while(pHead1 && pHead2){
            if(pHead1->val <= pHead2->val){
                p->next = pHead1;
                pHead1 = pHead1->next;
            }else{
                p->next = pHead2;
                pHead2 = pHead2->next;
            }
            p = p->next;
        }
	  //剩余节点直接链接到末尾
        if(pHead1){
            p->next = pHead1;
        }else if(pHead2){
            p->next = pHead2;
        }   
        return h;
    }
};