#力扣LeetCode面试题 02.02. 返回倒数第 k 个节点 @FDDLC

题目描述：
https://leetcode-cn.com/problems/kth-node-from-end-of-list-lcci/

Java代码一：

class Solution {
    public int getNodeSum(ListNode head){
        int sum=0;
        while(head!=null){
            head=head.next;
            sum++;
        }
        return sum;
    }
    public int kthToLast(ListNode head, int k) {
        int count=getNodeSum(head)-k;
        while(count-->0){
            head=head.next;
        }
        return head.val;
    }
}
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Java代码二：快慢指针

class Solution {
    public int kthToLast(ListNode head, int k) { //首末重合
        ListNode fast=head,slow=head;
        while(k-->0)fast=fast.next; //走k步
        if(fast==null)return head.val; //特判
        while(fast!=slow){
            slow=slow.next;
            fast=fast.next==null?head:fast.next;
            fast=fast.next==null?head:fast.next;
        }
        return fast.val; //slow.val亦可
    }
}
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Java代码三：等距同速指针

class Solution {
    public int kthToLast(ListNode head, int k) { //首末重合
        ListNode pioneer=head;
        while(k-->0)pioneer=pioneer.next;
        while(pioneer!=null){
            pioneer=pioneer.next;
            head=head.next;
        }
        return head.val;
    }
}
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注：有人把这种方法也叫做快慢指针，我认为是不妥的~