专访:合并两个排序单链表 [英] Interview: Merging two Sorted Singly Linked List
问题描述
这是一个编程问题接受记者采访时笔试时提出。 你有两个已排序单链表,你必须把它们合并,并返回新列表的头部,而无需创建任何新的额外的节点。返回的列表应该进行排序,以及
该方法的签名是: 节点MergeLists(List1的节点,节点list2中);
节点类如下:
类节点{
int数据;
下一节点;
}
我尝试了许多解决方案,但没有创造额外的节点螺丝的东西。请大家帮帮忙。
下面是附带的博客文章<一个href="http://techieme.in/merging-two-sorted-singly-linked-list/">http://techieme.in/merging-two-sorted-singly-linked-list/
节点MergeLists(List1的节点,节点list2中){
如果(List1的== NULL)返回list2中;
如果(list2中== NULL)返回List1中;
如果(list1.data&LT; list2.data){
list1.next = MergeLists(list1.next,list2中);
返回List1中;
} 其他 {
list2.next = MergeLists(list2.next,List1中);
返回list2中;
}
}
This is a programming question asked during a written test for an interview. "You have two singly linked lists that are already sorted, you have to merge them and return a the head of the new list without creating any new extra nodes. The returned list should be sorted as well"
The method signature is: Node MergeLists(Node list1, Node list2);
Node class is below:
class Node{
int data;
Node next;
}
I tried many solutions but not creating an extra node screws things. Please help.
Here is the accompanying blog entry http://techieme.in/merging-two-sorted-singly-linked-list/
Node MergeLists(Node list1, Node list2) {
if (list1 == null) return list2;
if (list2 == null) return list1;
if (list1.data < list2.data) {
list1.next = MergeLists(list1.next, list2);
return list1;
} else {
list2.next = MergeLists(list2.next, list1);
return list2;
}
}
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