[leetcode]Remove Nth Node From End of List

新博文地址：[leetcode]Remove Nth Node From End of List

 

http://oj.leetcode.com/problems/remove-nth-node-from-end-of-list/

 

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.

 虽然提示n始终合法，但是还是画蛇添足验证了一下：

边界情况有两种：

list == null

listLength=n【n = 1本来是特殊的，但是可以跟一般情况合并】

代码如下：

 public ListNode removeNthFromEnd(ListNode head, int n) {
    	int length = getLength(head);
    	if(head == null || n > length){
    		return null;
    	}
    	if(length == n){
   			return head.next;
    	}
    	ListNode tem1 = head;
    	ListNode tem2 = head;
    	//find the N+1th node from the end of list -- tem2's preNode
    	for(int i = 0 ; i < length; i++){
    		tem1 = tem1.next;    				
    		if(i > n){
    			tem2 = tem2.next;
    		}
    		if(tem1 == null){
    			break;
    		}
    	}
    	tem2.next = tem2.next.next;
    	
    	return head;
    }
    
    
    private int getLength(ListNode head){
    	int length = 0;
    	while(head != null){
    		head = head.next;
    		length++;
    	}
    	return length;
    }