题解 | #合并两个排序的链表#

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
            val(x), next(NULL) {
    }
};*/
class Solution {
  public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2) {

        ListNode* ans = new ListNode(-1);

        ListNode* cur = ans;

        while (pHead1 != nullptr && pHead2 != nullptr) {
            ListNode* tmp = new ListNode(-1);

            if (pHead1->val < pHead2->val) {
                tmp->val = pHead1->val;
                pHead1 = pHead1->next;
            } else {
                tmp->val = pHead2->val;
                pHead2 = pHead2->next;
            }

            cur->next = tmp;

            cur = cur->next;
        }

        // 注意这里  别忘 更新两个指针
        while (pHead1 != nullptr) {
            cur->next = pHead1;

            cur = cur->next;
            pHead1 = pHead1->next;
        }

        while (pHead2 != nullptr) {
            cur->next = pHead2;
			
            pHead2 = pHead2->next;
            cur = cur->next;
        }

        return ans->next == nullptr ? nullptr : ans->next;

    }
};

太容易大意