LeetCode：21_Merge Two Sorted Lists | 合並兩個排序列表 | Easy

題目：Merge Two Sorted Lists

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

簡單題，只要對兩個鏈表中的元素進行比較，然后移動即可，只要對鏈表的增刪操作熟悉，幾分鍾就可以寫出來，代碼如下：

struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x):val(x), next(NULL) {}
};

ListNode *GetLists(int n)    //得到一個列表
{
    ListNode *l = new ListNode(0);
    ListNode *pre = l;
    int val;
    for (int i = 0; i < n; i ++) {
        cin >> val;
        ListNode *newNode = new ListNode(val);
        pre->next = newNode;
        pre = pre->next;
    }
    return l->next;
}

ListNode *mergeTwoLists(ListNode *l1, ListNode *l2)
{
    assert (NULL != l1 && NULL != l2);
    if (NULL == l1 && NULL == l2)
        return NULL;
    if (NULL == l1 && NULL != l2) // !!要記得處理一個為空，另一個不為空的情況
        return l2;
    if (NULL != l1 && NULL == l2)
        return l1;
    
    ListNode *temp = new ListNode(0);
    temp->next = l1;
    ListNode *pre = temp;

    while(NULL != l1 && NULL != l2) {
        if (l1->val > l2->val) { //從小到大排列
            ListNode *next = l2->next;
            l2->next = pre->next;
            pre->next = l2;
            l2 = next;
        }        
        else {
            l1 = l1->next;
        }
        pre = pre->next;
    }
    if (NULL != l2) {
        pre->next = l2;
    }
    return temp->next;
}

這其中要注意一點，即要記得處理一個鏈表為空，另一個不為空的情況，如{}, {0} -- > {0},當然上面的寫法多少啰嗦了一些，可以簡寫。