leetcode 83. 删除排序链表中的重复元素

NOTE:

一、在下面文章中，给出了算法:

1、labuladong 如何高效对有序数组/链表去重？

2、labuladong 双指针技巧秒杀四道数组/链表题目

我的解答

#include <bits/stdc++.h>
using namespace std;

//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* removeDuplicates(ListNode *head, bool delete_flag = false)
    {
        if (head == nullptr)
        {
            return head;
        }
        ListNode *fast = head->next, *slow = head;
        while (fast)
        {
            if (fast->val != slow->val)
            {
                ListNode *delete_head = slow->next; // 待删除子链表的头
                slow->next = fast;
                slow = slow->next;
                /**
                 * 需要将两者之间的node全部都删除，因为它们都是重复的值
                 */
                if (delete_flag)
                {
                    while (delete_head != nullptr && delete_head != fast)
                    {
                        ListNode *n = delete_head;
                        delete_head = delete_head->next; // 迭代
                        cout << "删除:" << n->val << endl;
                        delete n; // 删除掉对应的节点
                    }
                }
            }
            fast = fast->next;
        }
        slow->next = nullptr;
        return head;
    }
};

ostream& operator <<(ostream &stream, ListNode *head)
{
    while (head)
    {
        stream << head->val << " ";
        head = head->next;
    }
    return stream;
}

void test1()
{
    ListNode *N1 = new ListNode { 1 };
    ListNode *N2 = new ListNode { 1 };
    ListNode *N3 = new ListNode { 2 };
    ListNode *N4 = new ListNode { 2 };
    ListNode *N5 = new ListNode { 5 };
    ListNode *N6 = new ListNode { 6 };
    N1->next = N2;
    N2->next = N3;
    N3->next = N4;
    N4->next = N5;
    N5->next = N6;
    cout << N1 << endl;
    Solution s;
    cout << s.removeDuplicates(N1, true) << endl;
}

void test2()
{
    ListNode N1 { 1 };
    ListNode N2 { 1 };
    ListNode N3 { 2 };
    ListNode N4 { 2 };
    ListNode N5 { 5 };
    ListNode N6 { 6 };
    N1.next = &N2;
    N2.next = &N3;
    N3.next = &N4;
    N4.next = &N5;
    N5.next = &N6;
    cout << &N1 << endl;
    Solution s;
    cout << s.removeDuplicates(&N1, false) << endl;
}
int main()
{
    test1();
    test2();
}
// g++ test.cpp -pedantic -Wall -Wextra --std=c++11