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19. 删除链表的倒数第N个节点.md

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60 lines (51 loc) · 1.11 KB

给定一个链表,删除链表的倒数第 n 个节点,并且返回链表的头结点。

示例:

给定一个链表: 1->2->3->4->5, 和 n = 2.

当删除了倒数第二个节点后,链表变为 1->2->3->5.

说明:

给定的 n 保证是有效的。

思路:

还是双指针,固定间隙为n

solution:

/**
 * 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) {
    if (head === null) return null
    const dum = {
        val: 0,
        next: head
    }
    let slow = dum
    let fast = head
    for (let i of Array.from({length: n})) {
        if (fast === null) {
            dum.next = dum.next.next
            return dum.next
        }
        fast = fast.next
    }
    function move (s, f) {
        if (f === null) {
            s.next = s.next.next
            return dum.next
        }
        s = s.next
        f = f.next
        return move(s, f)
    }
    return move(slow, fast)
};