Remove Nth Node From End of Listチェーンテーブル最後からN番目の要素@LeetCodeを削除

タイトル:
チェーンテーブル削除最後からN番目の要素
考え方:
最後からk番目の要素を見つけてから削除します.

package Level2;

import Utility.ListNode;

/**
 * 
 * Remove Nth Node From End of List 
 * 
 * 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.
 */
public class S19 {

	public static void main(String[] args) {
		ListNode head = new ListNode(1);
		ListNode l2 = new ListNode(2);
		ListNode l3 = new ListNode(3);
		ListNode l4 = new ListNode(4);
		ListNode l5 = new ListNode(5);
		head.next = l2;
		l2.next = l3;
		l3.next = l4;
		l4.next = l5;
		
		ListNode h = removeNthFromEnd(head, 5);
		h.print();
	}
	
	public static ListNode removeNthFromEnd(ListNode head, int n) {
		if(n == 0 || head == null){
			return head;
		}
		if(n == 1 && head.next==null){
			return null;
		}
		
		ListNode p = head, q = head;
		//  p  q n   
		for(int i=0; i<n; i++){
			if(p != null){
				p = p.next;
			}else{
				return head;
			}
		}
		
		//       p   null，         head
		if(p == null){
			head = head.next;
			return head;
		}
		
		// p q    
		while(p.next != null){
			q = q.next;
			p = p.next;
		}
		//     
		q.next = q.next.next;
		return head;
    }

}

Again:
チェーンテーブルの問題は必ずdummyHeadを使います!

/**
 * 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) {
        ListNode dm = new ListNode(0);
        dm.next = head;
        ListNode pro = dm, cur = dm;
        int i;
        for(i=0; i<n; i++){
            if(pro.next!=null){
                pro = pro.next;
            }else{
                break;
            }
        }
        if(i == n){
            while(pro.next != null){
                pro = pro.next;
                cur = cur.next;
            }
            if(cur.next != null){
                cur.next = cur.next.next;
            }
            return dm.next;
        }else{          // n too big to get
            return dm.next;
        }
    }
}