题解 | 合并两个排序的链表-遍历版

/**
 * 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* pDummy = new ListNode(-1);
        ListNode* pCurr = pDummy;

        while (pHead1 && pHead2) {
            if (pHead1->val < pHead2->val) {
                pCurr->next = pHead1;
                pHead1 = pHead1->next;
            }
            else {
                pCurr->next = pHead2;
                pHead2 = pHead2->next;
            }
            pCurr = pCurr->next;
        }
        pCurr->next = pHead1 ? pHead1 : pHead2;
        return pDummy->next;
    }
};