Remove Nth Node From End of List 链表删除倒数第N个元素@LeetCode

题目：

链表删除倒数第N个元素

思路：

先找到倒数第k个元素，再删除之

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;
// 让p先行q n个位置
for(int i=0; i<n; i++){
if(p != null){
p = p.next;
}else{
return head;
}
}

// 如果这个时候p已经是null，则说明删除的必定为head
if(p == null){
head = head.next;
return head;
}

// p和q一起前进
while(p.next != null){
q = q.next;
p = p.next;
}
// 删除元素
q.next = q.next.next;
return head;
}

}

Again:

链表题必用dummyHead！

/**
* 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;
}
}
}