链表24|19|面试题02.07|142

24两两交换链表中的节点

# 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]:
        dummy_head = ListNode()
        dummy_head.next = head
        cur = dummy_head
        while cur.next and cur.next.next:
            temp1 = cur.next
            cur.next = cur.next.next
            temp2 = cur.next.next
            cur.next.next = temp1
            temp1.next = temp2
            cur = cur.next.next
        return dummy_head.next

19删除链表的倒数第N个节点

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
        dummy_head = ListNode()
        dummy_head.next = head
        slow = dummy_head
        fast = dummy_head
        for i in range(n):
            fast = fast.next
        while fast.next:
            slow = slow.next
            fast = fast.next
        slow.next = slow.next.next
        return dummy_head.next

面试题02.07链表相交

# 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:
        curA = headA
        curB = headB
        sizeA = 0
        sizeB = 0
        while curA:
            sizeA += 1
            curA = curA.next
        while curB:
            sizeB += 1
            curB = curB.next
        curA = headA
        curB = headB
        if sizeA > sizeB:
            curA, curB = curB, curA
            sizeA, sizeB = sizeB, sizeA
        gap = sizeB - sizeA
        for i in range(gap):
            curB = curB.next
        while curA and curB:
            if curA == curB:
                return curA
            curA = curA.next
            curB = curB.next
        return None

142

# 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]:
        slow = head
        fast = head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if slow == fast:
                index1 = head
                index2 = fast
                while index1 != index2:
                    index1 = index1.next
                    index2 = index2.next
                return index1
        return None