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 ) {
    // write code here
    struct ListNode* head = NULL;
    struct ListNode* pre = NULL;
    int n = 0;
    int idx = 0;

    while (pHead1 != NULL || pHead2 != NULL) {
        struct ListNode* p = malloc(sizeof(struct ListNode));
        if (n == 0) {
            head = p;
            pre = p;
            n = 1;
        }
        if (pHead1 != NULL && pHead2 != NULL) {
            if (pHead1->val < pHead2->val) {
                p->val = pHead1->val;
                pHead1 = pHead1->next;
            } else {

                p->val = pHead2->val;
                pHead2 = pHead2->next;
            }
        } else if (pHead1 != NULL) {
            p->val = pHead1->val;
            pHead1 = pHead1->next;
        } else {
            p->val = pHead2->val;
            pHead2 = pHead2->next;
        }
        if (idx == 1) {
            pre ->next = p;
            pre = p;
        }
        idx = 1;
        p ->next = NULL;
    }

    return head;
}