[LeetCode][JavaScript]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.

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

 

 

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提示里说一遍循环就能搞定。

开个数组指向链表里的元素(js里存的是对象的引用)，数组的下标就是顺序，只要找倒数第n个数组元素，把他前一个指向他的后一个就好了。

异常边界值无非就是，没有前一个，没有后一个和链表只有一个元素。

我又要吐槽js好坑啊，空的话不应该返回{}吗， node节点都是对象啊，居然需要返回[]。

/**
 * @param {ListNode} head
 * @param {number} n
 * @return {ListNode}
 */
var removeNthFromEnd = function(head, n) {
    var cache = [];
    var i = 0;
    var current = head;
    while(current){
        cache[i] = current;
        i++;
        current = current.next;
    }

    var index = cache.length - n;
   
    if(!cache[index - 1]){
        return cache[1] || [];
    }else if(!cache[index + 1]){
        cache[index - 1].next = null;
        return cache[0]; 
    }else{
        cache[index - 1].next = cache[index + 1];
        return cache[0]; 
    }
};