leetCode 19.Remove Nth Node From End of List(删除倒数第n个节点) 解题思路和方法

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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.

思路：删除倒数第n个节点，因为是单链表，不知道节点总数，故先遍历，统计节点总数，算出正数第几个，然后删除即可。

算法很简单，代码如下：

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode removeNthFromEnd(ListNode head, int n) {        //删除倒数第n个        if(n == 0){            return head;        }        int nth = 0;//倒数第n个        int count = 0;//总的节点数        ListNode p = head;        //统计count        while(p != null){            p = p.next;            count++;        }        //计算正数n的值，从0计算        n = count - n;        if(n == 0){//如果为0，说明头结点，返回头结点下一个即可            return head.next;        }        p = head;        //数到n-1,然后令n-1.next = n.next = n-1.next.next即可        while(nth < n - 1){            p = p.next;            nth++;        }        p.next = p.next.next;        return head;    }}

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