• 0019. Remove Nth Node From End of List (M)


    Remove Nth Node From End of List (M)

    题目

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

    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.

    Follow up:

    Could you do this in one pass?


    题意

    给定一个链表,删除倒数第n个结点。

    思路

    One Pass方法:先将一个指针移动到正数第n个结点处,这时再建一个指针指向头结点,之后同步向后移动这两个指针。当第一个指针到达尾结点时,第二个指针正好在倒数第n个结点处。为了进行删除操作,还要建一个指针保存第二个指针的前结点,要注意删除的正好是头结点这种情况。


    代码实现

    Java

    /**
     * Definition for singly-linked list.
     * public class ListNode {
     *     int val;
     *     ListNode next;
     *     ListNode(int x) { val = x; }
     * }
     */
    class Solution {
        public ListNode removeNthFromEnd(ListNode head, int n) {
            int count = 1;
            ListNode last = head;
            while (count != n) {
                last = last.next;
                count++;
            }
            ListNode nth = head;
            ListNode prev = null;	// 只有当nth移动至少一次,prev才会指向具体的结点
            while (last.next != null) {
                last = last.next;
                prev = nth;
                nth = nth.next;
            }
            // 当prev仍为null,说明nth并没有移动,即要删除的就是头结点
            if (prev == null) {
                return nth.next;
            }
            prev.next = nth.next;
            nth.next = null;
            return head;
        }
    }
    

    JavaScript

    /**
     * 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) {
      let dummy = new ListNode(0, head)
      let p = dummy
      let count = 0
    
      while (count < n + 1) {
        p = p.next
        count++
      }
        
      let q = dummy
      while (p !== null) {
        p = p.next
        q = q.next
      }
      q.next = q.next.next
        
      return dummy.next
    }
    
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  • 原文地址:https://www.cnblogs.com/mapoos/p/13171276.html
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