How to Remove the Duplicates from Sorted List (Leaving Only Dist

  • 时间：2020-09-12 10:06:27
  • 分类：网络文摘
  • 阅读：148 次

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

Using a Hashmap to Count the Items

We can use a hashmap i.e. unordered_map in C++, to count the occurences of the items in the original linked list. This will cost O(N) time and O(N) space. However, the algorithm also applies to unsorted linked list. But it requires two passes.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        unordered_map<int, int> count;
        ListNode* p = head;
        while (p) {            
            count[p->val] ++;
            p = p->next;
        }
        ListNode* dummy = new ListNode(-1);
        p = dummy;
        while (head) {
            if (count[head->val] == 1) {
                p->next = new ListNode(head->val);
                p = p->next;
            }
            head = head->next;
        }
        return dummy->next;
    }
};
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        unordered_map<int, int> count;
        ListNode* p = head;
        while (p) {            
            count[p->val] ++;
            p = p->next;
        }
        ListNode* dummy = new ListNode(-1);
        p = dummy;
        while (head) {
            if (count[head->val] == 1) {
                p->next = new ListNode(head->val);
                p = p->next;
            }
            head = head->next;
        }
        return dummy->next;
    }
};

Skipping Duplicate Items on the Fly

The fact that the linked list is sorted helps us desgin a better algorithm. We can count the occurences of the current node in the linked list. If it is more than one, then skip it, otherwise, add the current node to the previous node – which we can update iteratedly.

This approach is O(N) time and O(1) constant space requirement.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        ListNode* dummy = new ListNode(-1);
        ListNode* prev = dummy;
        while (head) {
            int c = 0;
            ListNode *cur = head;
            while (head && head->val == cur->val) {
                c ++;
                head = head->next;
            }
            if (c == 1) {
                prev->next = cur;
                prev = cur;
            }
            cur->next = NULL;
        }        
        return dummy->next;
    }
};
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        ListNode* dummy = new ListNode(-1);
        ListNode* prev = dummy;
        while (head) {
            int c = 0;
            ListNode *cur = head;
            while (head && head->val == cur->val) {
                c ++;
                head = head->next;
            }
            if (c == 1) {
                prev->next = cur;
                prev = cur;
            }
            cur->next = NULL;
        }        
        return dummy->next;
    }
};

Depending on the requirements, you might want to delete the un-used nodes from the original linked list – in order to free the memory.

–EOF (The Ultimate Computing & Technology Blog) —

推荐阅读：
生活中的分数　 关于互质数的知识（五下）　 乘飞机带行李的尺寸规定　 位置问题的两种情况　 小九九和大九九　 加减乘除等号的来历　 什么是数据　 奇特的算法　 如何识别身份证号220103197102131101的真伪　 SEO优化和SEM搜索引擎营销，区别与联系全在这里了　
评论列表
添加评论