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  • 19. Remove Nth Node From End of List(js)

    19. Remove Nth Node From End of List

    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.
    题意:删除链表的倒数第n个节点
    代码如下:
    /**
     * 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.next==null) return null;
            //定义两个指针
            let pre=head,cur=head;
            //让cur沿着head向后移动n个节点
            for(let i=0;i<n;i++){
                cur=cur.next;
            }
            if(!cur) return head.next;
            //直至cur指向head末端
            while(cur.next){
                cur=cur.next;
                pre=pre.next;
            }
            pre.next=pre.next.next;
            return head; 
    };
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  • 原文地址:https://www.cnblogs.com/xingguozhiming/p/10387351.html
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