[LeetCode] 19. Remove Nth Node From End of List

题目：https://leetcode.com/problems/remove-nth-node-from-end-of-list/description/

题目

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.

思路

主要解决的问题，访问倒数的某个值很简单，但是要删去它，并且要将原链表链接起来。
C++技巧思路

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode* front = head;
ListNode** back = &head;    //对于 node->next ，设node1的next，为node2。在一般的想法中，如果一般想改变node1的back必须知道node1，然后访问给node1->next赋值，但是还有种思路就是对next本身赋值，就是要保存指向该值的指针，该指针为**p.
n--;
while(n){
n--;
front = front->next;
}
while(front->next!=NULL){
front = front->next;
back = &((*back)->next);
}
*back = (*back)->next;
return head;
}
};

Summary

This article is for beginners. It introduces the following idea: Linked List traversal and removal of nth element from the end.

Solution

Approach #1 (Two pass algorithm)

Intuition

We notice that the problem could be simply reduced to another one : Remove the (L
- n + 1)(L−n+1) th
node from the beginning in the list , where LL is
the list length. This problem is easy to solve once we found list length LL.

Algorithm

First we will add an auxiliary "dummy" node, which points to the list head. The "dummy" node is used to simplify some corner cases such as a list with only one node, or removing the head of the list. On the first pass, we find the list length LL.
Then we set a pointer to the dummy node and start to move it through the list till it comes to the (L
- n)(L−n) th
node. We relink 

next

 pointer of the (L
- n)(L−n) th
node to the (L
- n + 2)(L−n+2)th
node and we are done.



Figure 1. Remove the L - n + 1 th element from a list.

Java

/**
* 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 dummy = new ListNode(0);
dummy.next = head;
int length = 0;
ListNode first = head;
while(first!=null){
length++;
first = first.next;
}
length -= n;
first = dummy;
while(length>0){
length--;
first = first.next;
}
first.next = first.next.next;
return dummy.next;
}
}

Approach #2 (One pass algorithm)

Algorithm

The above algorithm could be optimized to one pass. Instead of one pointer, we could use two pointers. The first pointer advances the list by n+1n+1 steps
from the beginning, while the second pointer starts from the beginning of the list. Now, both pointers are exactly separated by nn nodes
apart. We maintain this constant gap by advancing both pointers together until the first pointer arrives past the last node. The second pointer will be pointing at the nnth
node counting from the last. We relink the next pointer of the node referenced by the second pointer to point to the node's next next node.



Figure 2. Remove the nth element from end of a list.

Java

/**
* 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 dummy = new ListNode(0);
dummy.next = head;
ListNode first = dummy;
ListNode second = dummy;
for(int i = 1;i<=n+1;i++){
first = first.next;
}
while(first!=null){
first =first.next;
second = second.next;
}
second.next = second.next.next;
return dummy.next;
}
}

note：构造 头结点 

1.规避了只有一个节点的情况。
2.操作时，second 节点位于 删去节点的上个节点，方便操作。