微软算法100题13 查找单向链表倒数第K个节点

第13 题：
题目：输入一个单向链表，输出该链表中倒数第k个结点。链表的倒数第0 个结点为链表的尾指针

思路：先遍历链表，求得总节点数N，则倒数第K个节点是正数N-K，时间复杂度为o(N+N-K)

package com.rui.microsoft;

public class Test13_FindKthNodeFromBack {

    public static void main(String[] args) {
        Node node1 = new Node(1);
        Node node2 = new Node(2);
        Node node3 = new Node(3);
        Node node4 = new Node(4);
        Node node5 = new Node(5);
        
        node1.next = node2;
        node2.next = node3;
        node3.next = node4;
        node4.next = node5;
        
        Node node = Test13_FindKthNodeFromBack.getKthFromBack(node1, 3);
        System.out.println(node.value);
    }
    
    public static Node getKthFromBack(Node head, int k){
        Node node = head;
        if(null == node)return null;
        int size = 0;
        while(null != node){
            node = node.next;
            size++;
        }
        
        int pos = size - k;
        if(pos < 0) return null;
        
        //put pointer to the head again
        int i = 0;
        node = head;
        while(i < pos){
            node = node.next;
            i++;
        }
        return node;
    }
    
    static class Node{
        int value;
        Node next;
        Node(int v){
            this.value = v;
        }
    }
}