[LeetCode] 82. Remove Duplicates from Sorted List II

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

Example 1:

Input: 1->2->3->3->4->4->5
Output: 1->2->5

Example 2:

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

Algorithm:

1. Insert a dummy node to the head of the input linked list for easier implementation since the root node may change.

2. Use a slow pointer to track the previous node with distinct number, a fast pointer to track the last node of the same duplicated number.

3. Find the last node of the same duplicated number. If this last node is right next to the previous node with distinct number we know this last node has a distinct number. Simply forward both pointer. Otherwise, we know this last node has some nodes before it that have the same value. Skip all these duplicated nodes and set the fast pointer to the next node and start a new duplication check for a new value.

 

/**
 * 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) {
        ListNode dummy = new ListNode(0), slow = dummy, fast = head;
        slow.next = fast;
        while(fast != null) {
            while (fast.next != null && fast.val == fast.next.val) {
                fast = fast.next;   
            }
            if (slow.next != fast) { 
                slow.next = fast.next; 
                fast = slow.next;     
            } else { 
                slow = slow.next;
                fast = fast.next;
            }       
        }
        return dummy.next;
    }
}

转载于:https://www.cnblogs.com/lz87/p/10460739.html