LeetCode - Remove Nth Node From End of List

 

        删除链表中倒数第n个节点，需要考虑的细节：链表为空时，链表只有一个节点时，n=1时。

 

     

/**
 * 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) {
        
        if(head == null || (head.next == null && n==1))
            return null;
        if(n == 0)
            return head;
        ListNode p = new ListNode(1), q = new ListNode(1);
        p = head;
        int count = 0;
        while(p != null) {
            p = p.next;
            count ++;
        }
        
        int cur = count - n;
        
        if(cur == 0) {
            ListNode ln = new ListNode(1);
            ln = head.next;
            head.next = null;
            return ln;
        }
        
        count = 0;
        p = head;
        while(p != null) {
            count ++;
            if(count == cur)
                break;
            p = p.next;
        }
        
        q = p.next;
        
        p.next = q.next;
        q.next = null;
        
        return head;
        
    }
}

 

转载于:https://www.cnblogs.com/wxisme/p/4388245.html