83. Remove Duplicates from Sorted List

题目：

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.

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

题解：

 单链表去重复。 node.next.val与当前node.val相同时移动node.next，不同则移动当前node。 Time Complexity - O(n)，Space Complexity - O(1)。

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

二刷：

方法和一刷一样。

Java:

Time Complexity - O(n)，Space Complexity - O(1)。

/**
 * 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 node = head;
        while (node != null && node.next != null) {
            if (node.val == node.next.val) {
                node.next = node.next.next;
            } else {
                node = node.next;
            }
        }
        return head;
    }
}

三刷:

Java:

/**
 * 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 node = head;

        while (node != null && node.next != null) {
            if (node.val == node.next.val) {
                node.next = node.next.next;
            } else {
                node = node.next;
            }
        }
        return head;
    }
}