83. Remove Duplicates from Sorted List(Easy)

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

Example 1:

Input: 1->1->2
Output: 1->2

Example 2:

Input: 1->1->2->3->3
Output: 1->2->3

solution

我的解法

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode deleteDuplicates(ListNode head){
        ListNode p = head , t = head;       //p指向当前结点,t指向去重后的尾结点
        while (p != null && p.next != null)
        {
            while (p.next != null && p.val == p.next.val)  //找到下一个结点值不重复的结点p.next
                p = p.next;
            if (p.next != null)   //将p.next连在t后面
            {
                p = p.next;
                t.next = p;
                t = p;
            }
            else                //如果p.next为null,说明去重已完成
                t.next = null;
        }
        return head;
    }
}

官方解法

public ListNode deleteDuplicates(ListNode head) {
    ListNode current = head;
    while (current != null && current.next != null) {
        if (current.next.val == current.val) {
            current.next = current.next.next;
        } else {
            current = current.next;
        }
    }
    return head;
}

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

总结

题意是给定一个有序链表，要求删掉里面重复的元素。

我的思路是用while循环遍历链表,接着用一个while循环找到与当前结点值不同的下一个结点,然后将找到的结点与当前新链表的尾结点连接起来,直到外层while循环结束.
官方思路简单粗暴,直接比较当前结点与下一结点的值,如果相等,则将当前结点与下一结点的下一结点连接起来,否则继续检查下一结点.

Notes
1.这种题感觉没必要想的很复杂,我就是考虑复杂了,导致我的解法比官方解法慢.
2.链表的插入与删除要特别注意头结点与尾结点.