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  • [leetcode] Remove Nth Node From End of List

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

    For 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.
    Try to do this in one pass.

    https://oj.leetcode.com/problems/remove-nth-node-from-end-of-list/

    思路1:双指针思想,两个指针相隔n-1,每次两个指针向后一步,当后面一个指针没有后继了,前面一个指针就是要删除的节点。注意:删除节点时需要删除指针的前驱pre;增加dummy head处理删除头节点的特殊情况。

    public class Solution {
    	public ListNode removeNthFromEnd(ListNode head, int n) {
    		if (head == null)
    			return null;
    		ListNode dummyHead = new ListNode(-1);
    		dummyHead.next = head;
    		ListNode s = head;
    		ListNode t = head;
    		n = n - 1;
    		for (int i = 0; i < n; i++) {
    			t = t.next;
    		}
    		ListNode preS = dummyHead;
    		while (t.next != null) {
    			preS = preS.next;
    			s = s.next;
    			t = t.next;
    		}
    		// remove s
    		preS.next = s.next;
    		s.next = null;
    		return dummyHead.next;
    
    	}
    
    	public static void main(String[] args) {
    		ListNode head = new ListNode(1);
    		head.next = new ListNode(2);
    		head.next.next = new ListNode(3);
    		head.next.next.next = new ListNode(4);
    		head.next.next.next.next = new ListNode(5);
    		ListNode newHead = new Solution().removeNthFromEnd(head, 1);
    
    	}
    
    
    
    }

    第二遍记录:

      注意head是null的情况。

      注意移动的条件是t.next!=null

    public class Solution {
        public ListNode removeNthFromEnd(ListNode head, int n) {
            if(head==null)
                return null;
            ListNode dummyHead = new ListNode(-1);
            dummyHead.next = head;
            n=n-1;
            ListNode t=head;
            for(int i=0;i<n;i++)
                t=t.next;
            ListNode pre =dummyHead;
            while(t.next!=null){
                pre=pre.next;
                t=t.next;
            }
            ListNode toRemove=pre.next;
            pre.next = toRemove.next;
            toRemove.next=null;
            
            return dummyHead.next;
        }
    }
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  • 原文地址:https://www.cnblogs.com/jdflyfly/p/3810697.html
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