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) {}
 * };
 */
#include <cstddef>
#include <cstdlib>
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* head = NULL;
        ListNode* end = NULL;
        ListNode* next = NULL;
        while(pHead1 != NULL && pHead2 != NULL){
            if(pHead1 -> val < pHead2 ->val) {
                next = pHead1 -> next;
                if(end == NULL) {
                    end = pHead1;
                }else {
                    end -> next = pHead1;
                    end = pHead1;
                }
                if(head == NULL) head = end;
                pHead1 = next;
            } else {
                next = pHead2 ->next;
                if(end == NULL) {
                    end = pHead2;
                } else {
                    end -> next = pHead2;
                    end = pHead2;
                }
                
                if(head == NULL) head = end;
                pHead2 = next;
            }
        }
        while(pHead1 != NULL) {
            next = pHead1 ->next;
            if(end == NULL) {
                end = pHead1;
            }else {
                end -> next = pHead1;
                end = pHead1;
            }
            if(head == NULL) head = end;
            pHead1 = next;
        }
        while (pHead2 != NULL) {
            next = pHead2 ->next;
            if(end == NULL) {
                end = pHead2;
            } else {
                end -> next = pHead2;
                end = pHead2;
            }
            if(head == NULL) head = end;
            pHead2 = next;
        }
        return head;
    }
};