牛客题霸NC53删除链表的倒数第n个节点Java题解

牛客题霸NC53删除链表的倒数第n个节点Java题解
https://www.nowcoder.com/practice/f95dcdafbde44b22a6d741baf71653f6?tpId=117&&tqId=34974&rp=1&ru=/ta/job-code-high&qru=/ta/job-code-high/question-ranking
方法：快慢指针
解题思路：利用快慢指针，让快指针former先走n步，然后再让快指针former和慢指针latter一起走，这样当快指针指向链表的末尾为null时，慢指针刚好指向链表的倒数第n个节点。再将第n个节点删除即可。这里利用了一个pre指针保存latter的前驱节点，便于删除操作。

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 * }
 */

public class Solution {
    /**
     * 
     * @param head ListNode类
     * @param n int整型
     * @return ListNode类
     */
    public ListNode removeNthFromEnd (ListNode head, int n) {
        if(head == null || n<=0){
            return null;
        }

        ListNode former = head;
        ListNode latter = head;

        ListNode pre = null;  //latter的前驱

        //former先走n步
        for(int i=0;i<n;i++){
            if(former == null){   //n大于链表长度时
                return null;
            }
            former = former.next;
        }
        //former和latter一起走
        while(former!=null){
            former = former.next;
            pre = latter;
            latter = latter.next;
        }

        //latter就是倒数第n个节点
        //删除latter节点
        if(pre!=null){
            pre.next = latter.next;
        }else{
            //当pre==null时，删除头节点 例如：{1}，1
            head = head.next;
        }
        return head;
    }
}