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) {
        ListNode* pre = new ListNode(0);
        ListNode* temp = nullptr;
        pre->next = pHead1;
        pHead1 = pre;
        while(pHead2)
        {
            if(pHead1->next==nullptr)
            {
                pHead1->next = pHead2;
                break;
            }
            else if(pHead2->val<=pHead1->next->val)
            {
                temp = pHead2->next;
                pHead2->next = pHead1->next;
                pHead1->next = pHead2;
                pHead2 = temp;
            }
            else {
                pHead1 = pHead1->next;
            }
        }
        return pre->next;
    }
};