Remove Duplicates from Sorted List II

描述

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.

中文：删除掉所有含有重复元素的结点。

分析：如果此元素与下一个元素是重复的，则删除此元素。

并且记录该元素是否重复。

根据上面的重复标志位，决定最后是否删除该元素。

class Solution {
public:
ListNode *deleteDuplicates(ListNode *head) {
if (head == nullptr) return head;
ListNode dummy(INT_MIN); //定义一个头结点
dummy.next = head;//让头结点指向链表第一个节点
ListNode *prev = &dummy, *cur = head;//定义两个指针结点
while (cur != nullptr) {
bool duplicated = false;
while (cur->next != nullptr && cur->val == cur->next->val) {
duplicated = true;
ListNode *temp = cur;
cur = cur->next;
delete temp;
}
if (duplicated) { // 根据重复位的标志，判断是否删除重复的最后一个元素
ListNode *temp = cur;
cur = cur->next;
delete temp;
continue;
}
prev->next = cur;
prev = prev->next;
cur = cur->next;
}
prev->next = cur;
return dummy.next;
}
};