19. Remove Nth Node From End of List.md

19. Remove Nth Node From End of List

Question:

Given a linked list, remove the n-th node from the end of list and return its head.

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.

Thinking:

• Method:

1. Iterate to get the length of the list.
2. Iterate again to remove
3. Trick: Use a dummy node to represent the head.

	/**
	 * 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) {
	        ListNode result = new ListNode(0);
	        result.next = head;
	        int len = 0;
	        ListNode temp = result;
	        while(temp.next != null){
	            temp = temp.next;
	            len ++;
	        }
	        if (len < n) return null;
	        int index = len + 1 - n;
	        len = 0;
	        temp = result;
	        while(temp.next != null){
	            len ++;
	            if(len == index){
	                temp.next = temp.next.next;//Remove
	                return result.next;
	            }
	            temp = temp.next;
	        }
	        return result.next;
	    }
	}

二刷

1. 快慢指针，先确定两个指针之间的距离。
2. 同时向后移动指针，定位出倒数第n个结点。
3. 通过改变next指针删除元素。

/**
 * 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) {
        ListNode fast = head, slow = head;
        for(int i = 0; i < n + 1; i++)
            fast = fast.next;
        while(fast != null){
            fast = fast.next;
            slow = slow.next;
        }
        slow.next = slow.next.next;
        return head;
    }
}