[LintCode] Remove Duplicates form Sorted List I & II

Remove Duplicates form Sorted List I

Problem

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

Example

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

Note

遇到后一个结点和当前结点相等，就用当前结点指向下一个的下一个结点。

Solution

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

Remove Duplicates form Sorted List II

Problem

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

Example

Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

Note

和上一道题不同的地方就是，需要用双指针操作，且最后要返回dummy.next，以防止head结点即duplicate的情况返回错误的结果。
令pre = dummy, cur = head，用cur进行查重操作，pre是cur的前结点。当cur和cur.next等值的时候，cur后移至第一个不等值的点，用pre指向新的cur即可。

Solution

public class Solution {
    public static ListNode deleteDuplicates(ListNode head) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = dummy, cur = head;
        while (cur != null && cur.next != null) {
            if (cur.val == cur.next.val) {
                int val = cur.val;
                while (cur != null && cur.val == val) cur = cur.next;
                pre.next = cur;
            }
            else {
                pre = pre.next;
                cur = cur.next;
            }
        }
        return dummy.next;
    }
}