題目描述:
給定一個鏈表,刪除鏈表的倒數第 n 個節點,並且返回鏈表的頭結點。
示例:
給定一個鏈表: 1->2->3->4->5, 和 n = 2.
當刪除了倒數第二個節點後,鏈表變爲 1->2->3->5.
代碼實現:
- 常規方法,第一次遍歷找到鏈表的長度,第二次刪除對應節點。
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @param {number} n
* @return {ListNode}
*/
var removeNthFromEnd = function(head, n) {
var preHead = new ListNode(null)
preHead.next = head
var len = 0
var first = head
while (first) {
len++
first = first.next
}
len -= n
first = preHead
while (len != 0) {
len--
first = first.next
}
first.next = first.next.next
return preHead.next
};
- 雙指針法,很精妙的解法,快指針先跑n,之後快慢一起跑,快指針跑到最後時慢指針對應的就是刪除的節點。
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @param {number} n
* @return {ListNode}
*/
var removeNthFromEnd = function(head, n) {
var preHead = new ListNode(null)
preHead.next = head
var fast = preHead
var slow = preHead
while (n != 0) {
fast = fast.next
n--
}
while (fast.next != null) {
fast = fast.next
slow = slow.next
}
slow.next = slow.next.next
return preHead.next
};