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5-1 链表,在节点间穿针引线 Reverse Linked List
题目: LeetCode 206. 反转链表
反转一个单链表。
示例:
输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL
进阶:
你可以迭代或递归地反转链表。你能否用两种方法解决这道题?
// 206. Reverse Linked List
// https://leetcode.com/problems/reverse-linked-list/description/
// 时间复杂度: O(n)
// 空间复杂度: O(1)
public class Solution1 {
// Definition for singly-linked list.
public class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}
public ListNode reverseList(ListNode head) {
ListNode pre = null;
ListNode cur = head;
while(cur != null){
ListNode next = cur.next;
cur.next = pre;
pre = cur;
cur = next;
}
return pre;
}
}
// 206. Reverse Linked List
// https://leetcode.com/problems/reverse-linked-list/description/
//
// 递归的方式反转链表
// 时间复杂度: O(n)
// 空间复杂度: O(n) - 注意,递归是占用空间的,占用空间的大小和递归深度成正比:)
public class Solution2 {
// Definition for singly-linked list.
public class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}
public ListNode reverseList(ListNode head) {
// 递归终止条件
if(head == null|| head.next == null)
return head;
ListNode rhead = reverseList(head.next);
// head->next此刻指向head后面的链表的尾节点
// head->next->next = head把head节点放在了尾部
head.next.next = head;
head.next = null;
return rhead;
}
}
5-2 测试你的链表程序
// 206. Reverse Linked List
// https://leetcode.com/problems/reverse-linked-list/description/
// 时间复杂度: O(n)
// 空间复杂度: O(1)
public class Solution {
public ListNode reverseList(ListNode head) {
ListNode pre = null;
ListNode cur = head;
while(cur != null){
ListNode next = cur.next;
cur.next = pre;
pre = cur;
cur = next;
}
return pre;
}
public static void main(String[] args) {
int[] nums = {1, 2, 3, 4, 5};
ListNode head = new ListNode(nums);
System.out.println(head);
ListNode head2 = (new Solution()).reverseList(head);
System.out.println(head2);
}
}
// Definition for singly-linked list.
// 在Java版本中,我们将LinkedList相关的测试辅助函数写在ListNode里
public class ListNode {
public int val;
public ListNode next = null;
public ListNode(int x) {
val = x;
}
// 根据n个元素的数组arr创建一个链表
// 使用arr为参数,创建另外一个ListNode的构造函数
public ListNode (int[] arr){
if(arr == null || arr.length == 0)
throw new IllegalArgumentException("arr can not be empty");
this.val = arr[0];
ListNode curNode = this;
for(int i = 1 ; i < arr.length ; i ++){
curNode.next = new ListNode(arr[i]);
curNode = curNode.next;
}
}
// 返回以当前ListNode为头结点的链表信息字符串
@Override
public String toString(){
StringBuilder s = new StringBuilder("");
ListNode curNode = this;
while(curNode != null){
s.append(Integer.toString(curNode.val));
s.append(" -> ");
curNode = curNode.next;
}
s.append("NULL");
return s.toString();
}
}