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

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;
    }
}

版权声明：本文为博主原创文章，未经博主允许不得转载。