Remove Nth Node From End of List leetcode java
2016-07-20 20:14
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视频讲解 http://v.youku.com/v_show/id_XMTY1MTMzNjAyNA==.html
(1)定义两个指针
ListNode fast = head;
ListNode slow = head;
(2)将快指针向前移动N步
(3.1)判断此时快指针是否已经到达尽头,如果是,头节点就是要删除的节点,返回head.next。
(3.2)将快慢两个指针同时以相同的速度往前移动,当快指针走到尽头的时候,慢指针的下一个位置就是倒数第N个节点,将慢指针next指向next.next.
(1)定义两个指针
ListNode fast = head;
ListNode slow = head;
(2)将快指针向前移动N步
(3.1)判断此时快指针是否已经到达尽头,如果是,头节点就是要删除的节点,返回head.next。
(3.2)将快慢两个指针同时以相同的速度往前移动,当快指针走到尽头的时候,慢指针的下一个位置就是倒数第N个节点,将慢指针next指向next.next.
public class Solution { public ListNode removeNthFromEnd(ListNode head, int n) { ListNode fast = head; ListNode slow = head; for(int i=0;i<n;i++){ fast = fast.next; } if(fast == null){ head = head.next; return head; } while(fast.next != null){ fast = fast.next; slow = slow.next; } slow.next = slow.next.next; return head; } }
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