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>0;k--){
            if(fast==NULL)
                return NULL;
            fast = fast->next;
        }
        while(fast!=NULL){
            fast = fast->next;
            slow = slow->next;
            }
        return slow;
        }
};