反转单链表和双链表。
题目
206. 反转链表 - 力扣(LeetCode)
https://leetcode.com/problems/reverse-linked-list
题目难度:Easy
解答1[Java]:迭代法/遍历法
核心思想
遍历一遍链表,遍历到下一个元素的时候,保留上一个元素的引用,这样就可以直接把当前元素的 next 设置为上一个元素。
每次遍历时候,先把下个节点临时保存下,然后把当前节点的 next 设置为 prev,然后再之前临时保存的下一个元素进行下一次遍历。
由于每次遍历的时候,都会使用下一个元素,当下一个元素变为 null 的时候,也就意味着遍历结束。
代码
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class Solution { public ListNode reverseList(ListNode head) { if (head == null) { return null; }
ListNode prev = null; ListNode current = head; while (current != null) { ListNode next = current.next; current.next = prev; prev = current; current = next; }
return prev; } }
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解答2[Java]:递归法
核心思想
先递归到最后一个节点,然后返回最后一个节点的指针。最后一个节点的指针一直传回递归的最顶层,同时每次递归返回之前,翻转当前节点和自己的下一个节点的关系。最后返回的指针是最后一个节点的指针,同时相邻节点的关系都被翻转了。
主要是利用 head.next.next = head;
相当于翻转了当前节点和下一个节点的关系。递归调用到最后,p 一直是指向最后一个节点,并且不断被返回到上一层调用的。
代码
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class Solution { public ListNode reverseList(ListNode head) { if (head == null || head.next == null) return head; ListNode p = reverseList(head.next); head.next.next = head; head.next = null; return p; } }
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拓展:反转双向链表
核心思路
代码
双向链表节点定义
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| public class DoubleNode { public int value; public DoubleNode last; public DoubleNode next;
public DoubleNode(int data) { this.value = data; } }
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算法
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| class Solution { public static DoubleNode reverseList(DoubleNode head) { DoubleNode pre = null; DoubleNode post = null; while (head != null) { post = head.next; head.next = pre; head.last = post; pre = head; head = post; } return pre; } }
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完整测试代码
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| public class Code_07_ReverseList { public static class Node { public int value; public Node next;
public Node(int data) { this.value = data; } }
public static Node reverseList(Node head) { Node pre = null; Node post = null; while (head != null) { post = head.next; head.next = pre; pre = head; head = post; } return pre; }
public static class DoubleNode { public int value; public DoubleNode last; public DoubleNode next;
public DoubleNode(int data) { this.value = data; } }
public static DoubleNode reverseList(DoubleNode head) { DoubleNode pre = null; DoubleNode post = null; while (head != null) { post = head.next; head.next = pre; head.last = post; pre = head; head = post; } return pre; }
public static void printLinkedList(Node head) { System.out.print("Linked List: "); while (head != null) { System.out.print(head.value + " "); head = head.next; } System.out.println(); }
public static void printDoubleLinkedList(DoubleNode head) { System.out.print("Double Linked List: "); DoubleNode end = null; while (head != null) { System.out.print(head.value + " "); end = head; head = head.next; } System.out.print("| "); while (end != null) { System.out.print(end.value + " "); end = end.last; } System.out.println(); }
public static void main(String[] args) { Node head1 = new Node(1); head1.next = new Node(2); head1.next.next = new Node(3); printLinkedList(head1); head1 = reverseList(head1); printLinkedList(head1);
DoubleNode head2 = new DoubleNode(1); head2.next = new DoubleNode(2); head2.next.last = head2; head2.next.next = new DoubleNode(3); head2.next.next.last = head2.next; head2.next.next.next = new DoubleNode(4); head2.next.next.next.last = head2.next.next; printDoubleLinkedList(head2); printDoubleLinkedList(reverseList(head2));
}
}
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相关题目
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实际考察
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补充题:反转双向链表
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小米二面
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参考资料