一起养成写作习惯！这是我参与「掘金日新计划 · 4 月更文挑战」的第14天，点击查看活动详情。
Given the head of a singly linked list, reverse the list, and return
the reversed list
.
Example 1:
Input: head = [1,2,3,4,5]
Output: [5,4,3,2,1]
Example 2:
Input: head = [1,2]
Output: [2,1]
Example 3:
Input: head = []
Output: []
Constraints:
- The number of nodes in the list is the range
[0, 5000]. -5000 <= Node.val <= 5000
Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both?
package com.offer;
/**
* @Author you guess
* @Date 2020/12/5 09:50
* @Version 1.0
* @Desc
*/
class ListNode {
int val;
ListNode next;
ListNode(int val, ListNode next) {
this.val = val;
this.next = next;
}
}
public class Leetcode_206_ReverseLinkedList {
/* static class ListNode {
int val;
ListNode next;
ListNode(int val, ListNode next) {
this.val = val;
this.next = next;
}
}*/
/**
* 更喜欢这个方法！！！
* 20220404
* <p>
* 方法1：cur起始指向第1个节点,prev起始指向null
* <p>
* Runtime: 0 ms, faster than 100.00% of Java online submissions for Reverse Linked List.
* Memory Usage: 44 MB, less than 6.23% of Java online submissions for Reverse Linked List.
*
* @param head
* @return
*/
public ListNode reverseListSolution(ListNode head) {
ListNode prev = null;
ListNode curr = head;
ListNode nextTemp = null;
while (curr != null) {
nextTemp = curr.next;
curr.next = prev;
prev = curr;
curr = nextTemp;
}
return prev;
}
/**
* 方法2：cur起始指向第二个节点，prev起始指向第一个节点
* <p>
* Runtime: 0 ms, faster than 100.00% of Java online submissions for Reverse Linked List.
* Memory Usage: 38.9 MB, less than 46.11% of Java online submissions for Reverse Linked List.
* <p>
* pre = cur.next;
* cur.next = pre;
* 以上2行是有区别的！！！！！
*
* @param head
* @return
*/
public ListNode reverseList(ListNode head) {
if (head == null)
return null;
ListNode pre = head;//指向第1个节点
ListNode cur = head.next;//指向第2个节点
ListNode post = null;
//ListNode post = head.next.next;//只有1个节点时报错NullPointerException，1->null，null.next报错NullPointerException
//此时pre是头指针，需要把next指针断掉，不然后面1<->2互相next会形成环，造成死循环；
//并且不能在while中写本行代码，不然pre往前的指向就断了！
pre.next = null;
while (cur != null) {
post = cur.next;//提前把下一个节点取到
cur.next = pre; //pre指针不变，变化的是cur.next，此时是要修改cur.next，所以不能写成pre=cur.next;
pre = cur; //cur指针不变，变化的是pre指针。pre,cur,post依次向前
cur = post;
}
return pre;
}
public void printListNode(ListNode head) {
while (head != null) {
System.out.println(head.val);
head = head.next;
}
}
public static void main(String[] args) {
ListNode node5 = new ListNode(5, null);
ListNode node4 = new ListNode(4, node5);
ListNode node3 = new ListNode(3, node4);
ListNode node2 = new ListNode(2, node3);
ListNode node1 = new ListNode(1, node2);
Leetcode_206_ReverseLinkedList main = new Leetcode_206_ReverseLinkedList();
main.printListNode(main.reverseList(node1));
// 输出：
// 5
// 4
// 3
// 2
// 1
}
}
end
