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 ) {
    struct ListNode* cur1 = pHead1;
    struct ListNode* cur2 = pHead2;
    struct ListNode* p = malloc(sizeof(struct ListNode));
    struct ListNode* head = p;

    if (pHead1 == NULL && pHead2 == NULL)   return NULL;
    if (pHead1 == NULL && pHead2 != NULL)   return pHead2;
    if (pHead1 != NULL && pHead2 == NULL)   return pHead1;

    while (cur1 != NULL && cur2 != NULL) {
        // 当pHead1的值小于等于pHead2的值时
        if (cur1->val <= cur2->val) {
            p->next = cur1;
            cur1 = cur1->next;
            p = p->next;
            if (cur1 == NULL) {
                p->next = cur2;
            }
        }
        // 当pHead1的值大于pHead2的值时
        else {
            p->next = cur2;
            cur2 = cur2->next;
            p = p->next;
            if (cur2 == NULL) {
                p->next = cur1;
            }
        }
    }
    return head->next;
}