[LeetCode][JavaScript]Remove Nth Node From End of List

標籤：

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

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.

 

 提示裡說一遍迴圈就能搞定。開個數組指向鏈表裡的元素(js裡存的是對象的引用)，數組的下標就是順序，只要找倒數第n個數組元素，把他前一個指向他的後一個就好了。異常邊界值無非就是，沒有前一個，沒有後一個和鏈表只有一個元素。我又要吐槽js好坑啊，空的話不應該返回{}嗎， node節點都是對象啊，他居然需要返回[]。 

 1 /** 2  * @param {ListNode} head 3  * @param {number} n 4  * @return {ListNode} 5  */ 6 var removeNthFromEnd = function(head, n) { 7     var cache = []; 8     var i = 0; 9     var current = head;10     while(current){11         cache[i] = current;12         i++;13         current = current.next;14     }15 16     var index = cache.length - n;17    18     if(!cache[index - 1]){19         return cache[1] || [];20     }else if(!cache[index + 1]){21         cache[index - 1].next = null;22         return cache[0]; 23     }else{24         cache[index - 1].next = cache[index + 1];25         return cache[0]; 26     }27 };

 

[LeetCode][JavaScript]Remove Nth Node From End of List