14链表中倒数第k个结点

题目描述

输入一个链表，输出该链表中倒数第k个结点。

 

思路：
快慢指针

快指针 先走k 步， 然后快慢指针一起走
当快指针走到null 时， 慢指针就是所求的倒数第ｋ个节点

ｔｉｐｓ：
判断ｋ是否越界是在快指针走的时候

 

 public class Solution {
     public ListNode FindKthToTail(ListNode head,int k) {
         // 边界条件
         if(head==null||head.next==null) return head;
         ListNode fast = head;
         ListNode slow = head;
         for(;k>0;k--){
             //如果k越界
             if(fast==null) return null;
             fast=fast.next;
         }
         while(fast!=null){
             fast=fast.next;
             slow=slow.next;
         }
         return slow;
     }
 }

 public class Solution {
     public ListNode FindKthToTail(ListNode head,int k) {
         if(head == null || head.next==null) return head;
         ListNode fast = head;
         ListNode slow = head;
         for(int i = 0;i<k;i++){
             if(fast==null) return null;
             fast = fast.next;
 
         }
         while(fast!= null){
             fast = fast.next;
             slow =slow.next;
         }
         return slow;
     }

20180307

 public class Solution {
     public ListNode FindKthToTail(ListNode head,int k) {
             if(head==null) return head;
             int cnt = 0;
             ListNode fast = head;
             ListNode slow = head;
             for(ListNode root = head;root != null;root = root.next)
                 cnt++;
             if(k>cnt||k<=0)return null;
             for(int i = 0;i<k;i++)
                 fast = fast.next;
             while(k!=0 && fast!=null){
                 fast = fast.next;
                 slow = slow.next;
             }
             return slow;
     }
 }

C++:20180808

 /*
 struct ListNode {
     int val;
     struct ListNode *next;
     ListNode(int x) :
             val(x), next(NULL) {
     }
 };*/
 class Solution {
 public:
     ListNode* FindKthToTail(ListNode* head, unsigned int k) {
         if(head==NULL) return NULL;
         ListNode* fast = head;
         ListNode* slow = head;
         for(;k>;k--){
             if(fast==NULL)
                 return NULL;
             fast = fast->next;
         }
         while(fast!=NULL){
             fast = fast->next;
             slow = slow->next;
             }
         return slow;
         }
 };