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[LeetCode][Medium] 19. Remove Nth Node From End of List

Given the head of a linked list, remove the nth node from the end of the list and return its head.

題目

Given the head of a linked list, remove the nth node from the end of the list and return its head.

A mapping of digit to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.

Example 1: alt

Input: head = [1,2,3,4,5], n = 2
Output: [1,2,3,5]

Example 2:

Input: head = [1], n = 1
Output: []

Example 3:

Input: head = [1,2], n = 1
Output: [1]

思路

用左右雙指針,先將右邊移動直到雙方相差 n,再讓左右同時移動直到右邊到底為止,這時左邊就會是後面數來第n個。

Solution

/**
 * Definition for singly-linked list.
 * 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) {
    var h = new ListNode(0)
    var l = h
    var r = h
    
    h.next = head
    
    for (var i = 0; i < n + 1; i++) {
        r = r.next
    }
    
    while(r !== null) {
        l = l.next
        r = r.next
    }
    
    l.next = l.next.next
    
    return h.next
};

Complexity

  • Time complexity : O(n)
  • Space complexity : O(1)

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