題目:Merge Two Sorted 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: 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} -- > ...
. 題目描述 Merge two sorted linked lists and return it as a newsortedlist. The new list should be made by splicing together the nodes of the first two lists. .示例 示例 : 示例 : 示例 : . 要求 The number of nodes i ...
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題目:Merge Two Sorted 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: 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} -- > ...
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 ...
題目: 給出兩個排序的單鏈表,合並兩個單鏈表,返回合並后的結果; 解題思路: 解法還是很簡單的,但是需要注意以下幾點: 1. 如果兩個鏈表都空,則返回null; 2. 如果鏈表1空,則返回鏈表2的頭節點;反之,如果鏈表2為空,則返回鏈表1的頭節點; 3. 兩個鏈表都不空的情況下 ...
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 ...
就是合並兩個有序鏈表了,遞歸解妥妥兒的。 ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) { if (l1 == NULL) return l2 ...
問題:有序合並兩個有序鏈表分析:歸並排序的合並部分 class Solution { public: ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) { ListNode *helper=new ...
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. Example:Input: 第二種方法的解決代碼 ...
的,之前做過一道 Merge Two Sorted Lists,是混合插入兩個有序鏈表。這道題增加了難度 ...