题解 | #合并两个排序的链表#
合并两个排序的链表
https://www.nowcoder.com/practice/d8b6b4358f774294a89de2a6ac4d9337
/** * struct ListNode { * int val; * struct ListNode *next; * ListNode(int x) : val(x), next(nullptr) {} * }; */ #include <cstddef> 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) { //如果一个为空的话直接返回另一个 if(pHead1==NULL) return pHead2; if(pHead2==NULL) return pHead1; // 创建新的链表表头 ListNode *head = new ListNode(0); ListNode *res = head; while(pHead1!=NULL && pHead2!=NULL) { if(pHead1->val <= pHead2->val) { res->next = pHead1; //移动取值的指针 pHead1 = pHead1->next; } else { res->next = pHead2; pHead2 = pHead2->next; } //指针向后移 res = res->next; } //哪个链表还有剩余直接连再后面 if(pHead1!=NULL) { res->next = pHead1; } else { res->next = pHead2; } return head->next; } };