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Remove Nth Node From End of List

时间:2015-09-12 16:06:50      阅读:150      评论:0      收藏:0      [点我收藏+]

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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:两个指针 p, q,其中q先走n步,之后如果q为null,则删除第一个node,否则,p, q开始一起走,直到q.next为null,此时p.next即是要删除的node.

JAVA代码:

public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode p = head;
        ListNode q = head;
        int i = 0;
        while (i < n) {
            q = q.next;
            i ++;
        }
        if (q == null) {
            head = head.next;
            return head;
        }
        while (q.next != null) {
            q = q.next;
            p = p.next;
        }
        p.next = p.next.next;
        return head;    
    }

 

Remove Nth Node From End of List

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原文地址:http://www.cnblogs.com/lasclocker/p/4803062.html

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