【LeetCode】Remove Duplicates from Sorted List 解题报告

Remove Duplicates from Sorted List

[LeetCode]

https://leetcode.com/problems/remove-duplicates-from-sorted-list/

Total Accepted: 114584 Total Submissions: 311665 Difficulty: Easy

Question

Given a sorted linked list, delete all duplicates such that each element appear only once.

For example,
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.

Ways

方法一

我的方法。嗯。比较智障。用了两个指针。一个指向现在正在看的这个元素叫做move，另一个指向后面的元素叫做next，每次比较move时，一次性的找出后面有几个和move相同的元素让next查找出来，让move跳过去。

确实智障。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if(head==null){
            return head;
        }
        ListNode next;
        ListNode move=head;
        while(move!=null&&move.next!=null){
             next=move.next;
            while(next!=null && move.val==next.val){
                next=next.next;
            }
            move.next=next;
            move=next;
        }

        return head;

    }
}

AC:2ms

方法二

LeetCode官方解答。

如果下一个元素和这个元素的值相等。这个元素的下个元素就等于下个元素的下个元素。再循环就好了。相当于只要下个跟我的相等，我就跳过你。

在找到不同的元素之前，当前元素不走。找到之后再走。

This is a simple problem that merely tests your ability to manipulate list node pointers. Because the input list is sorted, we can determine if a node is a duplicate by comparing its value to the node after it in the list. If it is a duplicate, we change the next pointer of the current node so that it skips the next node and points directly to the one after the next node.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        ListNode move=head;
        while(move!=null && move.next!=null){
            if(move.next.val == move.val){
                move.next=move.next.next;
            }else{
                move=move.next;
            }
        }

        return head;

    }
}

Date

2016/5/1 15:05:24