题目:
输入一个链表,反转链表后,输出新链表的表头。
答案:
解法一:
第一想法就是用pre存前一个,next存后一个,代码如下:
/*
public class ListNode {
int val;
ListNode next = null;
ListNode(int val) {
this.val = val;
}
}*/
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;
}
}
解法二:
递归,代码如下:
public class Solution {
public ListNode ReverseList(ListNode head) {
if(head==null||head.next==null){
return head;
}
ListNode node = ReverseList(head.next);
head.next.next = head;
head.next = null;
return node;
}
}