2.2.4 Remove Duplicates from Sorted List

来源：互联网发布：浙江外商直接投资数据编辑：程序博客网时间：2024/05/17 03:04

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

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.

我的思路：设两个指针pre, cur, 这道题有两种思路：

1 while cur.val == pre.val ，只更新cur；一旦不同，才更新pre.next = cur （遇到相同的都跳过，遇到不同的，才一次性把跳过的所有相同的都删掉）

2 没走一步，只要相同就删除： pre.next = cur.next

思路1：我开始的代码：

public class Solution {    public ListNode deleteDuplicates(ListNode head) {        if(head == null) return head;        ListNode pre = head;        ListNode cur = head;        while(cur.next != null){            if(cur.next.val == pre.val){                cur = cur.next;            }            else{                cur = cur.next;                pre.next = cur;                pre = pre.next;            }        }        return head;    }}

修改后的代码：只加了一行

Time: O(n), Space: O(1)

pre.next = cur.next;

public class Solution {    public ListNode deleteDuplicates(ListNode head) {        if(head == null) return head;        ListNode pre = head;        ListNode cur = head;        while(cur.next != null){            if(cur.next.val == pre.val){                cur = cur.next;            }            else{                cur = cur.next;                pre.next = cur;                pre = pre.next;            }        }        pre.next = cur.next;        return head;    }}

这种思路有个问题：如果cur.val != pre.val, 我们还是设pre.next = cur （实际上pre.next的指针没变，这步是多余的操作）

正确答案2：while嵌套while

public class Solution {    public ListNode deleteDuplicates(ListNode head) {        if(head == null) return null;        if(head.next == null) return head;        ListNode pre = head;        ListNode cur = head.next;        while(cur != null){            while(cur != null && cur.val == pre.val){                cur = cur.next;            }            pre.next = cur;            pre = cur;        }        return head;    }}

思路2：http://blog.csdn.net/linhuanmars/article/details/24354291

这种方法更好，没有多余操作

Time: O(n), Space: O(1)

public class Solution {    public ListNode deleteDuplicates(ListNode head) {        if(head == null) return head;        ListNode pre = head;        ListNode cur = head.next;        while(cur != null){            if(cur.val == pre.val){                pre.next = cur.next;            }            else{                pre = cur;            }            cur = cur.next;        }        return head;    }}

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