[Leetcode 19] 19 Remove Nth Node From End Of List

Problem:

Given a linked list, remove the n^th 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.

 

Analysis:

Use an auxilary pointer that n procedeed the current node. When the aux pointer is null, then the current pointer is the pointer to be removed;

Also need a pointer that 1 succeed the current point for update.

This is a one pass algorithm, so the time complecity is O(n); and space complexity is O(1);

 

Code:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        // Start typing your Java solution below
        // DO NOT write main() function
        ListNode toRmv = head, aux = head, preRmv = head;
        
        for (int i=0; i<n; ++i)
            aux = aux.next;
        
        while (aux != null) {
            preRmv = toRmv;
            toRmv = toRmv.next;
            aux = aux.next;
        }
        
        if (toRmv == head) {
            head = head.next;
        } else {
             preRmv.next = toRmv.next;
        }
        
        return head;
    }
}

 

 

Attention:

If the Node to be removed is the head, need process it separately.