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* head = new ListNode(0);
        auto raw_head = head;
        while (pHead1 != nullptr && pHead2 != nullptr) {
            if (pHead1->val < pHead2->val) {
                head->next = pHead1;
                pHead1 = pHead1->next;
            } else {
                head->next = pHead2;
                pHead2 = pHead2->next;
            }
            head = head->next;
        }
        if (pHead1 != nullptr) {
            head->next = pHead1;
        } else if(pHead2 != nullptr) {
            head->next = pHead2;
        }
        head = raw_head->next;
        delete raw_head;
        return head;
    }
};