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 == nullptr) return pHead2;
        if(pHead2 == nullptr) return pHead1;

         ListNode* dummy = new ListNode(0); //创建哨兵节点,防止未被delete就return了，放在判空之前，以防内存泄漏
        ListNode* cur = dummy;

        while (pHead1 != nullptr && pHead2 != nullptr) {
            if(pHead1->val <= pHead2->val)
            {
                cur->next = pHead1;
                pHead1 = pHead1->next;
            }
            else {
                cur->next = pHead2;
                pHead2 = pHead2->next;
            }
            cur = cur->next;
        }

        if(pHead1 != nullptr)
            cur->next = pHead1;
        else
            cur->next = pHead2;

        ListNode* result = dummy->next;
        delete dummy;
        return result;
        
    }
};