Leetcode | Remove Duplicates from Sorted List I && II

Remove Duplicates from Sorted List I

Given a sorted linked list, delete all duplicates such that each element appear only once.

For example,
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.

如果下一个节点和当前节点的值一样，就继续往后。这样找到的是重复数的最后一个。比如1->1->2; 

当然也可以找到重复数的第一个数；

注意delete掉那些重复的点。

class Solution {
public:
    ListNode *deleteDuplicates(ListNode *head) {
        ListNode *p = head, *newH = NULL, *tail, *tmp;
        while (p != NULL) {
            while (p->next != NULL && p->next->val == p->val) {
                tmp = p->next; 
                delete p;
                p = tmp;
            }
            if (newH == NULL) {
                newH = p;
            } else {
                tail->next = p; 
            }
            tail = p;
            p = p->next;
        }
        return newH;
    }
};

Remove Duplicates from Sorted List 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.

和Remove Duplicates from Sorted List I类似。加了一个判断是否重复数的条件。最后tail要指向NULL，因为原链表最后一个元素不一定会加进来。

这道题用c++很奇怪。下面的代码没有delete是可以accepted的。

class Solution {
public:
    ListNode *deleteDuplicates(ListNode *head) {
        ListNode *p = head, *newH = NULL, *tail, *current, *tmp;
        while (p != NULL) {
            current = p;
            while (p->next != NULL && p->next->val == p->val) p = p->next;
            if (p == current) {
                if (newH == NULL) {
                    newH = p;
                } else {
                    tail->next = p; 
                }
                tail = p;
            }
            p = p->next;
        }
        if (tail) tail->next = NULL;
        return newH;
    }
};

但是加了delete的逻辑之后就runtime error了。这份代码应该是没有问题的。难道是leetcode自己有问题？

class Solution {
public:
    ListNode *deleteDuplicates(ListNode *head) {
        ListNode *p = head, *newH = NULL, *tail, *current, *tmp;
        while (p != NULL) {
            current = p;
            while (p->next != NULL && p->next->val == p->val) {
                tmp = p->next; 
                delete p;
                p = tmp;
            }
            tmp = p->next;
            if (p == current) {
                if (newH == NULL) {
                    newH = p;
                } else {
                    tail->next = p; 
                }
                tail = p;
            } else {
                delete p;
            }
            p = tmp;
        }
        if (tail) tail->next = NULL;
        return newH;
    }
};