[LeetCode] 19. Remove Nth Node From End of List Java

Posted 2020-10-25 BrookLearnData

题目：Given a linked list, remove the nth node from the end of list and return its head.

For example,

   Given linked list: 1->2->3->4->5, and n = 2.
   After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.
Try to do this in one pass.

题意及分析：开始我的做法是先遍历一遍取得长度，然后长度减去n就知道要删除正着数第几个元素，不符合要求，因为只能遍历一次，所以只能用两个指针，一个快指针先走n步，一个慢指针从头开始走，这样当快指针走到尾部的时候，慢指针所指的就是要删除元素的前一个元素。

代码：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        if(head == null) return null;
        ListNode fast = head;//因为只能遍历一次，所以只能用两个指针，一个快指针先走n步，一个慢指针从头开始走，这样当快指针走到尾部的时候，慢指针所指的就是要删除元素的前一个元素。
        ListNode low = head;
        for(int i=0;i<n;i++){
            fast = fast.next;
        }
        if(fast == null){
            head = head.next;
            return head;
        }
        
        while(fast.next != null){
            low = low.next;
            fast = fast.next;
        }
        low.next = low.next.next;
        return head;
    }
}