真的是太久太久没有刷题了。。。。那天阿里面到这么简单的题目发现自己都写不利索了。。。哭瞎。。。
Reverse a singly linked list.
可以用递归和非递归的方法。
递归：
主要思想是定义一个nextNode，用nextNode作为尾部直接连前面一个。
public class Solution {
public ListNode reverseList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode nextNode = head.next;
ListNode newHead = reverseList(nextNode);
nextNode.next = head;
head.next = null;
return newHead;
}
}
非递归：
public class Solution {
public ListNode reverseList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode pre = null;
ListNode next = null;
while (head != null) {
next = head.next;
head.next = pre;
pre = head;
head = next;
}
return pre;
}
}
C++做法，虽然现在还不懂这个ListNode*是什么意思。。。。
递归：
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseList(ListNode* head) {
if (head == NULL || head -> next == NULL) {
return head;
}
ListNode* node = head->next;
ListNode* newHead = reverseList(node);
node -> next = head;
head -> next = NULL;
return newHead;
}
};
非递归：
class Solution {
public:
ListNode* reverseList(ListNode* head) {
if (head == NULL || head -> next == NULL) {
return head;
}
ListNode* pre = NULL;
while (head != NULL) {
ListNode* next = head -> next;
head -> next = pre;
pre = head;
head = next;
}
return pre;
}
};
原文：http://www.cnblogs.com/aprilyang/p/6943148.html