LeetCode: Remove Nth Node From End of List 解题报告

Remove Nth Node From End of List

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Question Solution 

 

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.

 

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SOLUTION 1:

1.使用快慢指针，快指针先行移动N步。用慢指针指向要移除的Node的前一个Node.

2. 使用dummy node作为head的前缀节点，这样就算是删除head也能轻松handle啦！

主页君是不是很聪明呀？ :)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        // 0017
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        
        ListNode slow = dummy;
        ListNode fast = dummy;
        
        // move fast N more than slow.
        while (n > 0) {
            fast = fast.next;
            // Bug 1: FORGET THE N--;
            n--;
        }
        
        while (fast.next != null) {
            fast = fast.next;
            slow = slow.next;
        }
        
        // Slow is the pre node of the node which we want to delete.
        slow.next = slow.next.next;
        
        return dummy.next;
    }
}

 

GITHUB (国内用户可能无法连接):

https://github.com/yuzhangcmu/LeetCode/blob/251766ffb832f2278f43a05e194ca76584bf14ea/list/RemoveNthFromEnd.java