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
 */
struct ListNode* Merge(struct ListNode* pHead1, struct ListNode* pHead2 ) {
    struct ListNode* P,* Q;
    struct ListNode* head = (struct ListNode*)malloc(sizeof(struct ListNode));
    head->next = NULL;
    Q = head;

    if(!pHead1) return pHead2;
    if(!pHead2) return pHead1;

    while(pHead1&&pHead2){
        if(pHead1->val <= pHead2->val){
            P = pHead1;
            pHead1 = pHead1->next;
        }else{
            P = pHead2;
            pHead2 = pHead2->next;
        }
        Q->next = P; // 后插法
        Q = P;
    }

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

    struct ListNode* nhead = head->next;
    free(head);
    return nhead;
}

不要忘记保存最后一个结点

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