[LeetCode] 83. Delete duplicate elements in the sorted list

题目

给定一个已排序的链表的头 head , 删除所有重复的元素,使每个元素只出现一次 .返回 已排序的链表 .

示例 1：

输入：head = [1,1,2]
输出：[1,2]

示例 2：

输入：head = [1,1,2,3,3]
输出：[1,2,3]

提示：

链表中节点数目在范围 [0, 300] 内
-100 <= Node.val <= 100
题目数据保证链表已经按升序 排列

题解

双指针,a pointer to the first occurrence of the previous digit,One points to the first occurrence of the next digit

/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode() : val(0), next(nullptr) {} * ListNode(int x) : val(x), next(nullptr) {} * ListNode(int x, ListNode *next) : val(x), next(next) {} * }; */
class Solution {
    
public:
    ListNode* deleteDuplicates(ListNode* head) {
    
        ListNode* temp = head;
        ListNode* temp1 = head;
        while(temp != nullptr)
        {
    
            while(temp!= nullptr && temp->val == temp1->val)
            {
    
                temp = temp->next;
            }
            temp1->next = temp;
            temp1 = temp1->next;
            
        }
        return head;
    }
};

一次循环

/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode() : val(0), next(nullptr) {} * ListNode(int x) : val(x), next(nullptr) {} * ListNode(int x, ListNode *next) : val(x), next(next) {} * }; */
class Solution {
    
public:
    ListNode* deleteDuplicates(ListNode* head) {
    
        if(head==nullptr || head->next==nullptr)
            return head;
        
        ListNode* temp = head;
        while(temp->next)
        {
    
            if(temp->val == temp->next->val)
                temp->next = temp->next->next;
            else
                temp = temp->next;
        }
        return head;
    }
};