LeetCode19:Remove Nth Node From End of List

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.

/** * 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        int dist = 0;        ListNode left=head, right=head;        while(right.next!=null){            if(dist<n){                right=right.next;                dist++;            }            else{                right=right.next;                left=left.next;            }         }        if(left==head && dist<n)            head = head.next;        else            left.next = left.next.next;        return head;    }}

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LL's solution:

/** * 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 demi = new ListNode(0);        demi.next = head;        ListNode current = demi;        ListNode end = demi;                //move end to the Nth node from head        while(end.next!=null && n>0){            end = end.next;            n--;        }                if(n==0){            //move current and end concurrently until end reach the end of list            while(end.next!=null){                end = end.next;                current = current.next;            }                        //remove current            current.next = current.next.next;        }                return demi.next;    }}

Note:

Use a demi node to handle 'removing head' situation.