LeetCode 19：Remove Nth Node From End of List

   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.

一个链表，给出一个n，删除倒序后第n个节点，然后返回删除后的链表，比较简单

我这里的做法，先遍历整个链表，得到length，然后根据length和n得到要删除节点的位置，删除即可

有两种情况不要遗漏了，一种是n比length还大，另一种n和length相等

Java源码

package com.maoxiaomeng;

/**
 * @author lihui
 * @date 2018/3/26 12:31
 */

class ListNode { 
    int val;
    ListNode next;
    ListNode(int x) { 
        val = x; 
    }
}

public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode p = head;
        
        int len = 0;
        while (p != null) {
            len++;
            p = p.next;
        }
        
        int index = len - n;
        ListNode pStr = head;
        if (index < 0) {
            return p;
        } else if (index == 0) {
            return pStr.next;
        }
        int i = 0;
        /**
         *  0->1->2->3->4->5->6->7->8->9
         *                       i
         */
        while (pStr != null) {
            if (i == index - 1) {
                pStr.next = pStr.next.next;
                break;
            }
            pStr = pStr.next;
            i++;
        }
        return head;
    }
}

一次到位