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) {
        ListNode* res = nullptr;
        //边界情况
        if (pHead1 == nullptr || pHead2 == nullptr) {
            if(pHead1 == nullptr && pHead2 != nullptr) return pHead2;
            if(pHead1 != nullptr && pHead2 == nullptr) return pHead1;
            if(pHead1 == nullptr && pHead2 == nullptr) return res;
        }
        //尾插最大值
        ListNode* newnode = new ListNode(1001);
        ListNode* newnode1 = new ListNode(1001);
        ListNode* p1 = pHead1;
        ListNode* p2 = pHead2;
        while (p1->next != nullptr) p1 = p1->next;
        while (p2->next != nullptr) p2 = p2->next;
        p1->next = newnode;
        p2->next = newnode1;
        //
        while (pHead1->next != nullptr || pHead2->next != nullptr) {
            if (pHead1->val < pHead2->val) {
                ListNode* newnode = new ListNode(pHead1->val);
                pHead1 = pHead1->next;
                if (res == nullptr) {
                    res = newnode;
                } else {
                    ListNode* ptemp = res;
                    while (ptemp->next != nullptr) ptemp = ptemp->next;
                    ptemp -> next = newnode;
                }
            } else if (pHead1->val > pHead2->val) {
                ListNode* newnode = new ListNode(pHead2->val);
                pHead2 = pHead2->next;
                if (res == nullptr) {
                    res = newnode;
                } else {
                    ListNode* ptemp = res;
                    while (ptemp->next != nullptr) ptemp = ptemp->next;
                    ptemp -> next = newnode;
                }
            } else {
                ListNode* newnode = new ListNode(pHead2->val);
                ListNode* newnode1 = new ListNode(pHead2->val);
                newnode->next = newnode1;
                pHead1 = pHead1->next;
                pHead2 = pHead2->next;
                if (res == nullptr) {
                    res = newnode;
                } else {
                    ListNode* ptemp = res;
                    while (ptemp->next != nullptr) ptemp = ptemp->next;
                    ptemp -> next = newnode;
                }
            }
        }
        return res;
    }
};

尾插最大值

不断地尾插