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  • 19. Remove Nth Node From End of List

    https://leetcode.com/problemset/all/?search=19

    涉及链表删除操作的时候,稳妥起见都用 dummy
    node,省去很多麻烦。因为不一定什么时候 head
    被删了。

    /**
     * Definition for singly-linked list.
     * public class ListNode {
     *     int val;
     *     ListNode next;
     *     ListNode(int x) { val = x; }
     * }
     */
    public class Solution {
        public ListNode removeNthFromEnd(ListNode head, int n) {
            if (head == null) {
                return head;
            }
            int len = 0;
            ListNode dummy = new ListNode(0);
            dummy.next = head;
            while (head != null) {
                len++;
                head = head.next;
            }
            head = dummy;
            int index = len - n;
            while (index > 0) {
                head = head.next;
                index--;
            }
            head.next = head.next.next;
            return dummy.next;
        }
    }
    

     快慢指针

    public ListNode removeNthFromEnd(ListNode head, int n) {
        
        ListNode start = new ListNode(0);
        ListNode slow = start, fast = start;
        slow.next = head;
        
        //Move fast in front so that the gap between slow and fast becomes n
        for(int i=1; i<=n+1; i++)   {
            fast = fast.next;
        }
        //Move fast to the end, maintaining the gap
        while(fast != null) {
            slow = slow.next;
            fast = fast.next;
        }
        //Skip the desired node
        slow.next = slow.next.next;
        return start.next;
    }
    

      

     

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  • 原文地址:https://www.cnblogs.com/apanda009/p/7103122.html
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