代码随想录算法训练营Day04｜链表

24. 两两交换链表中的节点
使用交换的写法

# 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_node = ListNode(0,head)
        cur = dummy_node
        while cur.next and cur.next.next:
            temp = cur.next
            temp1 = cur.next.next.next

            cur.next = cur.next.next
            cur.next.next = temp
            cur.next.next.next = temp1
            cur = cur.next.next
        return dummy_node.next

使用并行交换的写法：

class Solution:
    def swapPairs(self, head: Optional[ListNode]) -> Optional[ListNode]:
        dummy_node = ListNode(0,head)
        cur = dummy_node
        while cur.next and cur.next.next:
            temp = cur.next
            temp1 = cur.next.next.next

            cur.next,cur.next.next, = cur.next.next,temp,
            cur = cur.next.next
            cur.next = temp1
        return dummy_node.next

19

# 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_node = ListNode(0,head)
        slow,fast = dummy_node,dummy_node
        for _ in range(n):
            fast = fast.next
        while fast.next:
            fast = fast.next
            slow = slow.next
        slow.next = slow.next.next
        return dummy_node.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_node = ListNode(0,head)
        slow,fast = dummy_node,dummy_node
        for _ in range(n):
            fast = fast.next
        while fast.next:
            fast = fast.next
            slow = slow.next
        slow.next = slow.next.next
        return dummy_node.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:
        lenA, lenB = 0,0
        cur = headA
        # 求出两个链表长度
        while cur:
            cur = cur.next
            lenA += 1
        cur = headB
        while cur:
            cur = cur.next
            lenB += 1
        # 择出长链表表头
        curA,curB = headA,headB
        if lenA > lenB:
            curA, curB = curB, curA
            lenA, lenB = lenB, lenA
        # 末位对齐
        for _ in range(lenB-lenA):
            curB = curB.next
        while curA:
            if curA == curB:
                return curA
            else:
                curA = curA.next
                curB = curB.next
        return None

142. 环形链表 II

# 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,fast = head,head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if slow == fast:
                slow = head
                # 再走一个z
                while slow != fast:
                    slow = slow.next
                    fast = fast.next
                return slow
        return None