题解 | #合并两个排序的链表#

/**
 * struct ListNode {
 *  int val;
 *  struct ListNode *next;
 * };
 */
/**
 * 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
 */
#include <math.h>
#include <stdlib.h>
struct ListNode* Merge(struct ListNode* pHead1, struct ListNode* pHead2) {
    // write code here
    struct ListNode* pHead =  (struct ListNode*)malloc(sizeof(struct ListNode));
    struct ListNode* head = pHead;
    struct ListNode* pre = pHead;
    // 检查是否有链表为空
    if(pHead1 == NULL)
    {
        return pHead2;
    }
    if(pHead2 == NULL) {
        return pHead1;
    }
    // 使用两个指针分别指向两个链表，进行值比较
    while (pHead1 != NULL && pHead2 != NULL) {
        if (pHead1->val < pHead2->val) {
            pHead->val = pHead1->val;
            struct ListNode* temp = (struct ListNode*)malloc(sizeof(struct ListNode));
            pHead->next = temp;
            pre = pHead;  // pre指针指向当前节点的上一个节点
            pHead = pHead->next;
            pHead1 = pHead1->next;
        } else {
            pHead->val = pHead2->val;
            struct ListNode* temp = (struct ListNode*)malloc(sizeof(struct ListNode));
            pHead->next = temp;
            pre = pHead;
            pHead = pHead->next;
            pHead2 = pHead2->next;
        }
    }
    if(pHead1 == NULL)
    {
         pre->next = pHead2;  // 接上另一个没遍历完的后续系节点
    }
    else {
         pre->next = pHead1;
    }
    return head;
}