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)
1 package com.rui.microsoft; 2 3 public class Test13_FindKthNodeFromBack { 4 5 public static void main(String[] args) { 6 Node node1 = new Node(1); 7 Node node2 = new Node(2); 8 Node node3 = new Node(3); 9 Node node4 = new Node(4); 10 Node node5 = new Node(5); 11 12 node1.next = node2; 13 node2.next = node3; 14 node3.next = node4; 15 node4.next = node5; 16 17 Node node = Test13_FindKthNodeFromBack.getKthFromBack(node1, 3); 18 System.out.println(node.value); 19 } 20 21 public static Node getKthFromBack(Node head, int k){ 22 Node node = head; 23 if(null == node)return null; 24 int size = 0; 25 while(null != node){ 26 node = node.next; 27 size++; 28 } 29 30 int pos = size - k; 31 if(pos < 0) return null; 32 33 //put pointer to the head again 34 int i = 0; 35 node = head; 36 while(i < pos){ 37 node = node.next; 38 i++; 39 } 40 return node; 41 } 42 43 static class Node{ 44 int value; 45 Node next; 46 Node(int v){ 47 this.value = v; 48 } 49 } 50 }
