[LeetCode] Remove Duplicates from Sorted List、Remove Duplicates from Sorted List II

最新推荐文章于 2023-08-30 08:07:21 发布

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本文介绍如何从已排序的链表中删除重复元素，包括仅删除出现一次的重复项及删除所有重复节点的情况。提供了两种算法实现，一种保留唯一元素，另一种则移除所有重复元素。

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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.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
#define LN ListNode
class Solution {
public:
    ListNode *deleteDuplicates(ListNode *head) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function

        const int INVALID =INT_MAX;
	    LN guard(INVALID);
	    guard.next=head;
	    LN* pre=&guard;
	    LN* cur=head;
	    while(cur)
	    {
		    if ( cur->val == pre->val )
		    {
			    pre->next=cur->next;
			    delete cur;
			    cur=pre->next;
		    }
		    else
		    {
			    pre=pre->next;
			    cur=cur->next;
		    }
	    }
	    return guard.next;
    }
};

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.

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

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
#define LN ListNode
class Solution {
public:
    ListNode *deleteDuplicates(ListNode *head) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        const int INVALID=INT_MAX;
	    LN guard(INVALID);
	    guard.next=head;
	    
	    LN* pTail=&guard;
	    LN* cur=head;
	    while(cur)
	    {
		    if ( cur->next==NULL || cur->val!=cur->next->val)
		    {
			    pTail->next=cur;
			    pTail=pTail->next;
			    cur=cur->next;
		    }
		    else
		    {
			    int v=cur->val;
			    while(cur&&cur->val==v)
			    {
				    pTail->next=cur->next;
				    delete cur;
				    cur=pTail->next;
			    }
		    }
	    }
	    return guard.next;
    }
};