[LeetCode]87. Remove Duplicates from Sorted List II排序链表去重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.

 

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解法1：一个简单的想法是使用map<int,int>来保存链表中每个不同元素出现的次数，然后遍历map，只要key对应的val不为1，则删除链表中所有值为key的节点。

/**
 * 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) {
        map<int, int> m;
        ListNode* curr = head;
        while (curr != NULL) {
            ++m[curr->val];
            curr = curr->next;
        }
        ListNode* help = new ListNode(0);
        help->next = head;
        curr = help;
        map<int, int>::iterator iter = m.begin();
        for (; iter != m.end(); ++iter) {
            if (iter->second != 1) {
                while (curr->next != NULL && curr->next->val != iter->first)
                    curr = curr->next;
                for (int i = 0; i < iter->second; ++i) {
                    ListNode* del = curr->next;
                    curr->next = curr->next->next;
                    delete del;
                }
            }
        }
        return help->next;
    }
};

 

解法2：可以降低空间复杂度为O(1)。如果找到相邻的两个节点值一样，则同时删除这两个节点，并且记录下这个值，如果接下来的节点还是这个值也删除即可。

/**
 * 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) {
        if (head == NULL || head->next == NULL) return head;
        ListNode* help = new ListNode(0);
        help->next = head;
        ListNode *first = help, *second = head, *third = head->next;
        while (third != NULL) {
            if (second->val == third->val) {
                int dupVal = second->val;
                while (second != NULL && second->val == dupVal) {
                    ListNode* del = second;
                    first->next = second->next;
                    second = second->next;
                    delete del;
                }
                if (second == NULL) break;
                third = second->next;
            }
            else {
                first = second;
                second = third;
                third = third->next;
            }
        }
        return help->next;
    }
};