leetcode 19. Remove Nth Node From End of List

最新推荐文章于 2024-01-12 15:01:11 发布

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最新推荐文章于 2024-01-12 15:01:11 发布

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本文介绍了一种高效算法，用于从链表中删除倒数第N个节点，并返回修改后的链表头。该算法使用了两种方法：递归实现单次遍历和双指针技巧。递归方法通过计数来定位目标节点，而双指针方法则通过保持两个指针间的距离来找到正确位置。

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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.

采用递归的方法实现one pass

public class Solution {
    int cnt=0, target=0;
    public ListNode removeNthFromEnd(ListNode head, int n) {
        target = n;
        return removeNthFromEnd(head);
    }
    private ListNode removeNthFromEnd(ListNode head){
        if(head==null){
            cnt++;
            return null; 
        }
        head.next = removeNthFromEnd(head.next);
        return cnt++==target? head.next : head;
    }
}

利用两个指针实现的方法

A one pass solution can be done using pointers. Move one pointer fast --> n+1 places forward, to maintain a gap of n between the two pointers and then move both at the same speed. Finally, when the fast pointer reaches the end, the slow pointer will be n+1 places behind - just the right spot for it to be able to skip the next node.

Since the question gives that n is valid, not too many checks have to be put in place. Otherwise, this would be necessary.

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;
}