[英]How to determine the space and time complexity of these two double linked list algorithms?
我解决了下一个练习,有两个解决方案: https : //www.hackerrank.com/challenges/reverse-a-doubly-linked-list
首先(非递归):
/*
Insert Node at the end of a linked list
head pointer input could be NULL as well for empty list
Node is defined as
class Node {
int data;
Node next;
Node prev;
}
*/
Node Reverse(Node head) {
if (head == null) return null;
Node current = head;
while (current.next != null) {
Node temp = current.next;
current.next = current.prev;
current.prev = temp;
current = temp;
}
current.next = current.prev;
current.prev = null;
return current;
}
第二种算法(递归):
/*
Insert Node at the end of a linked list
head pointer input could be NULL as well for empty list
Node is defined as
class Node {
int data;
Node next;
Node prev;
}
*/
Node Reverse(Node head) {
if (head.next == null) {
head.next = head.prev;
head.prev = null;
return head;
}
Node newHead = Reverse(head.next);
Node temp = head.next;
head.next = head.prev;
head.prev = temp;
return newHead;
}
根据这本书,解决方案必须是 O(n)。 我想使用递归解决方案更优雅,但也许我错了。 你能帮忙确定这两种算法的空间和时间复杂度,或者你的,哪个性能更好?
问题有点不清楚,两个解决方案在时间和空间上似乎都是 O(n)。 尽管您可能会删除特殊情况并使 Torvalds 高兴。 就像是:
Node Reverse(Node head) {
if (head == null) return null;
Node current = head;
while (current != null) {
Node temp = current.next;
current.next = current.prev;
current.prev = temp;
current = temp;
}
return current;
}
Node Reverse(Node head) {
Node temp = head.next;
head.next = head.prev;
head.prev = temp;
return temp==null?head:Reverse(temp);
}
我没有测试过这些,仅将它们用作灵感。 (如果 head 在开始时为空,则递归也会为空指针)。
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