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 <stdlib.h>
struct ListNode* Merge(struct ListNode* pHead1, struct ListNode* pHead2 ) {
    if (pHead1 == NULL) return pHead2;
    if (pHead2 == NULL) return pHead1;

    struct ListNode* returnData = NULL;
    struct ListNode* tail = NULL; // 用于追踪链表末尾

    while (pHead1 != NULL && pHead2 != NULL) {
        struct ListNode* newNode = (struct ListNode*)malloc(sizeof(struct ListNode));

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

        newNode->next = NULL;

        if (returnData == NULL) {
            returnData = newNode;
            tail = returnData;
        } else {
            tail->next = newNode;
            tail = tail->next;
        }
    }

    if (pHead1 != NULL) {
        tail->next = pHead1;
    } else if (pHead2 != NULL) {
        tail->next = pHead2;
    }

    return returnData;
}