1、取链表后k个元素/第k个元素
思路:
两个指针,快的先走k步,之后快慢指针一起走,当快指针到达尾部,慢指针也到达第k个节点了
2、取链表的中间节点
思路:
两个指针,一个一次走一步,一个一次走两步,当后一个到达尾部时,第一个指针即指向中间节点
代码:
package com.datastructure.link;
/*
* 数据结构之单链表
*/
public class SingleLinkedList2
{
/*
* 元素的基本定义
*/
static class Node{
/*
* value
*/
private int value;
/*
* 链表的下一个元素
*/
private Node next;
public Node (int value) {
this.value = value;
}
@Override
public String toString() {
if (this.next == null) {
return String.valueOf(this.value);
}
return this.value + "->" + this.next.toString();
}
}
/**
* 获取链表后k个元素/获取链表倒数第K个元素
*/
public static Node getLastKEle(Node node, int k) {
if (node == null || k <= 0) {
System.out.println("Invalid param");
return null;
}
Node fast = node;
Node later = node;
// 快的指针先走k步
for (int i = 0; i < k; i++) {
fast = fast.next;
if (fast == null) {
// 链表长度不足k,返回整个链表
return node;
}
}
// 继续一起遍历
while (fast != null) {
later = later.next;
fast = fast.next;
}
return later;
}
/**
* 两个指针 一个一次走一步 一个一次走两步
* 此函数,当链表元素为偶数时,取中间两个的后一个元素
*/
public static Node findMidNode(Node node) {
if (node == null) {
return null;
}
Node first = node;
Node second = node;
while (second != null && second.next != null) {
first = first.next;
second = second.next.next;
}
return first;
}
/**
* 两个指针 一个一次走一步 一个一次走两步
* 此函数,当链表元素为偶数时,取中间两个的前一个元素
*/
public static Node findMidPreNode(Node node) {
if (node == null) {
return null;
}
Node first = node;
Node second = node;
while (second != null && second.next != null && second.next.next != null) {
first = first.next;
second = second.next.next;
}
return first;
}
public static void main(String args[]) {
// 原链表
System.out.println("Origin link: " + createTestLinkedList(9));
// 打印链表后2个元素
System.out.println("print last 2 data: " + getLastKEle(createTestLinkedList(9), 2));
System.out.println("find mid last one: " + findMidNode(createTestLinkedList(8)));
System.out.println("find mid pre one: " + findMidPreNode(createTestLinkedList(8)));
}
private static Node createTestLinkedList(int n) {
Node head = new Node(0);
Node curNode = head;
for (int i = 1; i < n; i++) {
curNode.next = new Node(i);
curNode = curNode.next;
}
return head;
}
}