【Leetcode】 83. Remove Duplicates from Sorted List 【两个指针】

来源：互联网发布：sql server 2005 端口编辑：程序博客网时间：2024/05/16 01:38

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.

这道题的要求是在有序链表中删除有重复数字的节点，使其只出现1次。

思路就是找到重复元素，删除即可。循环检测，如果两个连续节点的数字一样，删除一个重复的元素。

时间复杂度：O(n)

空间复杂度：O(1)

代码：

/** * 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 || head.next == null)  return head;        ListNode begin = head;        ListNode curr = head.next;        ListNode last = head;        while( curr != null ){            if(last.val == curr.val){                last.next = curr.next;                curr.next = null;                curr = last.next;            }            else{                last = curr;                curr = curr.next;            }        }        return begin;    }}

注意：while循环的写法

另一种做法：（并没有卵用，除了能记录一下重复的次数，再删掉）

/** * 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 || head.next == null) return head;        int step = 0 ;        ListNode begin = head ;        ListNode curr = head.next;        ListNode last = head;                while(curr != null){            if(curr.val == last.val){                step ++;                curr = curr.next;                if(curr == null){                    if(step != 0){                        last.next = curr;                    }                }            }            else{                if(step == 0){                    last = curr;                    curr = curr.next;                }                else if(step != 0){                    last.next = curr;                    last = curr;                    curr = curr.next;                    step = 0 ;                }            }        }        return begin;    }}

0 0