删除且保留其中一个重复元素的思路和删除且不保留其中一个重复元素的思路

思路&＃xff1a;
遍历结点&＃xff0c;遇到相同的跳过一个结点&＃xff08;删除&＃xff09;

def deleteDuplicates(self, head):if head &＃61;&＃61; None:return headcur &＃61; headwhile cur.next !&＃61; None:first &＃61; cur.valccur &＃61; cur.nextsecond &＃61; ccur.val# 删除重复结点&＃xff0c;保留其中一个if second &＃61;&＃61; first:ccur &＃61; ccur.nextcur.next &＃61; ccurelse:cur &＃61; ccurreturn head

删除且不保留其中一个重复元素

思路&＃xff1a;
双指针&＃xff1a;快慢指针&＃xff0c;用快指针找到相同的或者不同的值&＃xff0c;并用flag标记&＃xff0c;找到不同的值后用慢指针指向。

def deleteDuplicates(self, head: ListNode) -> ListNode:dummy &＃61; ListNode(0)slow &＃61; dummyfast &＃61; headwhile fast !&＃61; None:# 是否快进的标志flag &＃61; Falsewhile fast.next !&＃61;None and fast.val &＃61;&＃61; fast.next.val:flag &＃61; Truefast &＃61; fast.nextif not flag:slow.next &＃61; fastslow &＃61; slow.nextfast &＃61; fast.nextslow.next &＃61; Nonereturn dummy.next