链表倒数第K个元素与链表的中间节点(Java)

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;
    }

}

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转载自blog.csdn.net/zangdaiyang1991/article/details/88584807
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