LeetCode 19:Remove Nth Node From End of List

作者: LiHui 分类: LeetCode 发布时间: 2018-04-12 15:08

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.

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

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