[LeetCode] 19. Remove Nth Node From End of List ☆
2017-02-18 17:02
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Given a linked list, remove the nth node from the end of list and return its head.
For example,
Note:
Given n will always be valid.
Try to do this in one pass.
解法:
题目要求移除倒数第n个节点,并且说明了n一定是有效的,限定一次遍历解决问题。首先定义两个指针 first 和 last,first 指向第一个节点,last 指向第 n+1 个节点;两个节点同时向后撸,直到 last.next 为 null 的时候,说明当前 first.next 即为待删除节点,直接 first.next.next = first.next 即可删除,然后返回 head。注意,如果一开始的时候 last 就为 null,说明 head 节点就是带删除的节点,直接返回 head.next 即可。
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.
解法:
题目要求移除倒数第n个节点,并且说明了n一定是有效的,限定一次遍历解决问题。首先定义两个指针 first 和 last,first 指向第一个节点,last 指向第 n+1 个节点;两个节点同时向后撸,直到 last.next 为 null 的时候,说明当前 first.next 即为待删除节点,直接 first.next.next = first.next 即可删除,然后返回 head。注意,如果一开始的时候 last 就为 null,说明 head 节点就是带删除的节点,直接返回 head.next 即可。
/** * 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) { ListNode first = head; ListNode last = head; while (n-- > 0) { last = last.next; } if (last == null) { return head.next; } while (last.next != null) { first = first.next; last = last.next; } first.next = first.next.next; return head; } }
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