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
        if (!pHead1) {
            return pHead2;
        } 
        if (!pHead2) {
            return pHead1;
        }
        auto* pre = new ListNode(-1);
        auto* new_head = pre;
        pre->next = pHead2;

        ListNode* p1 = pHead1;
        ListNode* p2 = pHead2;
        while (p1 && p2) {
            if (p1->val < p2->val) {
                pre->next = p1;
                p1 = p1->next;
            } else {
                pre->next = p2;
                p2 = p2->next;
            }
            pre = pre->next;
        }
        if (p1) {
            pre->next = p1;
        }
        if (p2) {
            pre->next = p2;
        }

        return new_head->next;
    }
};