Java for LeetCode 019 Remove Nth Node From End of List

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Given a linked list, remove the n^th 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.

解题思路：

难度不大，注意下边界条件即可：

JAVA实现：

static public ListNode removeNthFromEnd(ListNode head, int n) {
		if(n<=0)
			return head;
		ListNode ln=head;
		int i=1;
		while(ln.next!=null){
			ln=ln.next;
			i++;
		}
		if(i==n)
			return head.next;
		ln=head;
		for(;i>n+1;i--)
			ln=ln.next;
		ln.next=ln.next.next;
		return head;
	}

Java for LeetCode 019 Remove Nth Node From End of List

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原文地址：http://www.cnblogs.com/tonyluis/p/4473394.html