力扣算法题—083删除排序链表中的重复数字

给定一个排序链表，删除所有重复的元素，使得每个元素只出现一次。

示例 1:

输入: 1->1->2
输出: 1->2

示例 2:

输入: 1->1->2->3->3
输出: 1->2->3

#include "_000库函数.h"

//这道题与上题的不同之处就是重复的数字不用全部删除，保留一个

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

//有释放的删除代码
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 (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;
            }
            p = p->next;
        }
        return head->next;
    }
};


//无释放的删除节点
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;
            p = p->next;
        }
        return head->next;
    }
};


//使用递归
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if (!head || !head->next) return head;
        head->next = deleteDuplicates(head->next);
        return (head->val == head->next->val) ? head->next : head;
    }
};

void T083() {
    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;
}