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) {
        if(pHead1==NULL)
        {
            return pHead2;
        }
        if(pHead2 == NULL)
        {
            return pHead1;
        }
        // write code here
        ListNode *head = NULL;
        if(pHead1->val < pHead2->val)
        {
            head = pHead1;
            pHead1 = pHead1->next;
        }
        else 
        {
            head = pHead2;
            pHead2 = pHead2->next;
        }
        ListNode *cur = head;
        while(pHead1 && pHead2)
        {
            if(pHead1->val<pHead2->val)
            {
                cur->next = pHead1;
                cur = pHead1;
                pHead1 = pHead1->next;
            }
            else {
                cur->next = pHead2;
                cur = pHead2;
                pHead2 = pHead2->next;
            }
        }
        if(pHead1)
        {
            cur->next = pHead1;
        }
        if(pHead2)
        {
            cur->next = pHead2;
        }
        return head;
    }
};