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
 */

typedef struct ListNode ListNode;
struct ListNode* Merge(struct ListNode* pHead1, struct ListNode* pHead2 ) {
    // write code here
    ListNode* HeadNode, *TailNode;
    HeadNode = TailNode = (ListNode*)malloc(sizeof(ListNode));
    ListNode* p1 = pHead1;
    ListNode* p2 = pHead2;
    while (p1 && p2) {
        if (p1->val < p2->val) {
            TailNode->next = p1;
            TailNode = TailNode->next;
            p1 = p1->next;
        } else {
            TailNode->next = p2;
            TailNode = TailNode->next;
            p2 = p2->next;
        }
    }
    if (p1) {
        TailNode->next = p1;
    }
    if (p2) {
        TailNode->next = p2;
    }
    return HeadNode->next;
}