[LeetCode79]Remove Duplicates from Sorted List

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本文介绍了一种从已排序链表中删除重复元素的方法，使得每个元素仅出现一次。通过双指针技巧实现，当两个相邻节点值相等时跳过重复节点，不相等则正常遍历。提供了 C++ 和 Java 两种语言的实现代码。

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.

Analysis:

The idea is to use two pointer to scan, compare and remove the duplicates from the start to end.

if p1.val == p2.val, p2 = p2.next

else p1.next = p2, p1 = p1.next, p2 = p2.next

c++

ListNode *deleteDuplicates(ListNode *head) {
    if(head==NULL||head->next==NULL) return head;
    ListNode *p1 = head;
    ListNode *p2 = head->next;
    while(p2!=NULL){
        if(p1->val == p2->val){
            p2 = p2->next;
        }
        else{
            p1->next = p2;
            p2 = p2->next;
            p1 = p1->next;
        }
    }
    p1->next = p2;
    return head;
}

Java

public ListNode deleteDuplicates(ListNode head) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
       ListNode n1=head;
       while(n1 != null && n1.next !=null){
    	   if(n1.val == n1.next.val){
    		   n1.next = n1.next.next;
    	   }
    	   else
    		   n1 = n1.next;
       }
       return head;
    }