题目: 给你一个链表,删除链表的倒数第 n个结点,并且返回链表的头结点。题目链接
我的JavaScript解法:
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} head
* @param {number} n
* @return {ListNode}
*/
var removeNthFromEnd = function(head, n) {
const dummy = new ListNode(-1);
dummy.next = head;
let x = findKthFromEnd(dummy, n+1)
x.next = x.next.next;
return dummy.next
};
var findKthFromEnd = function (head, k) {
let p1 = head;
for(let i=0; i< k; i++) {
p1 = p1.next;
}
let p2 = head;
while(p1!=null) {
p2 = p2.next;
p1 = p1.next;
}
return p2;
};
解析: 快慢指针问题
- 时间复杂度:
- 空间复杂度: