leetcode

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.

题目的意思是将重复的结点全部删除，只留下不重复的结点，并且结点是排好序的

个人思路：

1，从链表头部开始遍历，设置3个指针，一个是current，表示当前结点，另一个是currentBefore，表示当前结点的上一个结点，最后一个是probe，表示当前结点的下一个结点，probe用于探索下面的结点是否和当前结点相同；

2，若相同，则删除probe结点，并且继续往下探索，直到probe结点和current结点不同，此时还得删除current结点，然后重新设置好这3个指针；若不同，则重新设置好这3个指针，继续往下遍历即可

3，注意一下头节点的处理

代码：

#include <stddef.h>

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

class Solution {
public:
    ListNode *deleteDuplicates(ListNode *head) {
        if (!head)
        {
            return NULL;
        }

        ListNode *current = head;
        ListNode *currentBefore = NULL;
        ListNode *probe = NULL;
        bool isDeleted;

        while (current)
        {
            isDeleted = false;
            probe = current->next;
            while (probe && probe->val == current->val) //删除相同的结点
            {
                current->next = probe->next;
                delete probe;
                probe = current->next;
                isDeleted = true;
            }
            if (isDeleted)
            {
                if (current == head)
                {
                    delete current;
                    current = probe;
                    head = current;
                }
                else
                {
                    currentBefore->next = probe;
                    delete current;
                    current = currentBefore->next;
                }
            }
            else
            {
                currentBefore = current;
                current = probe;
            }
        }

        return head;
    }
};