力扣算法题—082删除排序链表中的重复元素2

给定一个排序链表，删除所有含有重复数字的节点，只保留原始链表中 没有重复出现 的数字。

示例 1:

输入: 1->2->3->3->4->4->5
输出: 1->2->5

示例 2:

输入: 1->1->1->2->3
输出: 2->3

#include "_000库函数.h"


struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};


//又是无头结点的链表！！！
//还有，哪个数字重复了，就将他全部删除！！
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        //创建一个头结点
        ListNode *p = new ListNode(0);
        p->next = head;
        head = p;
        ListNode *q = p;
        int flag = 0;//用来标记重复数字
        while (q && p) {
            q = q->next;
            while (q && q->next && q->val == q->next->val) {
                //一定的记得释放删除的数字
                ListNode *ptr = q;
                p->next = q->next;
                delete(ptr);
                ptr = NULL;
                q = p->next;
                flag = 1;
            }
            if (flag) {//删除最后一个重复了的数字
                ListNode *ptr = q;
                p->next = q->next;
                delete(ptr);
                ptr = NULL;
                q = p;
                flag = 0;
            }
            else
                p = p->next;
        }
        return head->next;
    }
};

//不释放删除的节点，我不建议，但好多博客是这样写的
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        //创建一个头结点
        ListNode *p = new ListNode(0);
        p->next = head;
        head = p;
        ListNode *q = p;
        while (p->next) {
            q = p->next;
            while (q->next && q->val == q->next->val)
                q = q->next;
            if (p->next != q)
                p->next = q->next;
            else
                p = p->next;
        }
        return head->next;
    }
};



//使用递归，不使用while

class Solution {
public:
    ListNode *deleteDuplicates(ListNode *head) {
        if (!head) return head;
        if (head->next && head->val == head->next->val) {
            while (head->next && head->val == head->next->val) {
                head = head->next;
            }
            return deleteDuplicates(head->next);
        }
        head->next = deleteDuplicates(head->next);
        return head;
    }
};
void T082() {
    ListNode *head = new ListNode(0);
    ListNode *p = head;
    vector<int>v = { 1,2,3,4,4, 5, 5};
    for (auto a : v) {
        ListNode *q = new ListNode(0);
        q->val = a;
        p->next = q;
        p = q;
    }
    p = head->next;
    while (p) {
        cout << p->val << "->";
        p = p->next;
    }
    cout << endl;
    Solution s;
    p = s.deleteDuplicates(head->next);
    while (p) {
        cout << p->val << "->";
        p = p->next;
    }
    cout << endl;
}