Leetcode刷题第十六天-链表
24：两两交换链表中的节点
链接：24. 两两交换链表中的节点 - 力扣（LeetCode）
虚拟头节点
# Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def swapPairs(self, head: Optional[ListNode]) -> Optional[ListNode]: if not head or not head.next: return head dummpy=ListNode(next=head) left,cur=dummpy,head while cur and cur.next: right=cur.next tmp=right.next left.next=right right.next=cur cur.next=tmp left=cur cur=cur.next return dummpy.next
19:删除链表的倒数第n个节点
链接：19. 删除链表的倒数第 N 个结点 - 力扣（LeetCode）
# Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]: if not head: return head dummpy=ListNode(next=head) cur=dummpy.next lens=0 while cur: cur=cur.next lens+=1 cur=dummpy for i in range(lens-n): cur=cur.next if(cur.next): cur.next=cur.next.next return dummpy.next
链表相交
链接：面试题 02.07. 链表相交 - 力扣（LeetCode）
# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode: lenA,lenB=self.get_len(headA),self.get_len(headB) if(lenA>lenB): for i in range(lenA-lenB): headA=headA.next else: for i in range(lenB-lenA): headB=headB.next while headA: if(headA==headB): return headA headA=headA.next headB=headB.next return None def get_len(self,head): lens=0 while head: lens+=1 head=head.next return lens
142:环形链表II
链接：142. 环形链表 II - 力扣（LeetCode）
快指针一次走两步，慢指针一次走一步，快=慢，有环：慢指针从头开始走，快指针一次走一步，快=慢，环入口
# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]: if(not head): return head fast,slow=head,head while fast and fast.next: fast=fast.next.next slow=slow.next if fast==slow: slow=head while slow!=fast: slow=slow.next fast=fast.next return slow return None
2:两数相加
# Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: cur1,cur2=l1,l2 head=ListNode() re,cur=head,head num=0 while cur1 or cur2: if(cur1): c1=cur1.val cur1=cur1.next else: c1=0 if(cur2): c2=cur2.val cur2=cur2.next else: c2=0 sums=c1+c2+num if(sums<10): re.val=sums num=0 else: num=1 re.val=sums%10 if(cur1 or cur2): re.next=ListNode() re=re.next if(num==1): re.next=ListNode(1) return head
21:合并两个有序链表
链接：21. 合并两个有序链表 - 力扣（LeetCode）
# Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]: if(not list1): return list2 if(not list2): return list1 if(list1.val<list2.val): re=ListNode(list1.val) list1=list1.next else: re=ListNode(list2.val) list2=list2.next cur=re while list1 or list2: if(not list1): cur.next=list2 break if(not list2): cur.next=list1 break if(list1.val<list2.val): val=list1.val list1=list1.next else: val=list2.val list2=list2.next cur.next=ListNode(val) cur=cur.next return re
23:合并k个升序链表
链接：23. 合并 K 个升序链表 - 力扣（LeetCode）
链表转list，list转链表
# Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]: re=[] for node in lists: while node: re.append(node.val) node=node.next re.sort() if(len(re)==0): return None head=ListNode(re[0]) cur=head for i in range(1,len(re)): cur.next=ListNode(re[i]) cur=cur.next return head
25:K个一组翻转链表
链接：25. K 个一组翻转链表 - 力扣（LeetCode）
cur记录翻转前一个位置，left翻转起点，right翻转终点
# Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]: if(not head): return head dummpy=ListNode(next=head) cur,left,right=dummpy,head,dummpy i=0 while right: if(i==k): tmp=right.next right.next=None cur.next=self.reverseList(left) left.next=tmp right,cur=left,left left=left.next i=0 i+=1 right=right.next return dummpy.next def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]: pre,cur=None,head while cur: tmp=cur.next cur.next=pre pre,cur=cur,tmp return pre
61:旋转链表
k可能会大于链表长度，直接对链表长度取余，加个虚拟头节点，找到倒数lens-k-1的位置，头指向这个位置，链表尾指向head
# Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def rotateRight(self, head: Optional[ListNode], k: int) -> Optional[ListNode]: if(not head): return head dummpy=ListNode(next=head) cur=head lens=0 while cur: lens+=1 cur=cur.next n=k%lens right=head for i in range(lens-n-1): right=right.next dummpy.next=right.next right.next=None cur=dummpy while cur.next: cur=cur.next cur.next=head return dummpy.next
82:删除排序链表中的重复元素II
链接：82. 删除排序链表中的重复元素 II - 力扣（LeetCode）
# Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]: dummpy=ListNode(next=head) cur,pre=head,dummpy while cur and cur.next: if(cur.val==cur.next.val): val=cur.val while cur and cur.val==val: cur=cur.next pre.next=cur cur=pre.next else: pre=cur cur=cur.next return dummpy.next
83:删除排序链表中的重复元素
链接：83. 删除排序链表中的重复元素 - 力扣（LeetCode）
# Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]: uset=set() cur,pre=head,head while cur: if(cur.val not in uset): uset.add(cur.val) pre=cur cur=cur.next else: pre.next=cur.next cur=pre.next return head
