有环单链表相交判断、单链表相交判断

有环单链表相交判断 单链表相交判断

有环单链表相交判断 如何判断两个有环单链表是否相交？相交的话返回第一个相交的节点，不相交的话返回空。如果两个链表长度分别为N和M，请做到时间复杂度O(N+M)，额外空间复杂度O(1)。 给定两个链表的头结点head1和head2(注意，另外两个参数adjust0和adjust1用于调整数据,与本题求解无关)。请返回一个bool值代表它们是否相交。 我的提交 （这题意描述的不清不楚啊！！！没看懂题意：到底要返回啥？？？）

-- coding:utf-8 --

class ListNode:

def init(self, x):

self.val = x

self.next = None

class ChkIntersection: def chkInter(self, head1, head2, adjust0, adjust1): # write code here loopa, lena = self.chkLoop(head1) loopb, lenb = self.chkLoop(head2)

    if loopa != loopb:
        pa = loopa
        pb = loopb
        while pa != pb and pa != loopa:
            if pa == pb:
                # 两个入环点都在环内
                return pb
            pa = pa.next
        return None

    else:
        pa, pb = None
        if lena > lenb:
            pa, pb = head1, head2
        else:
            pa, pb = head2, head1

        for _ in range(abs(lena - lenb)):
            pa = pa.next

        while pa != pb:
            pa = pa.next
            pb = pb.next
        # 链表在入环前或入环时相交
        return pa

def chkLoop(self, head):
    # 返回有环链表的入环结点和链表长度
    slow = head
    fast = head
    n = 1
    while True:
        slow = slow.next
        n += 1
        fast = fast.next.next
        if fast == slow:
            break

    p = head
    while p != slow:
        p = p.next
        slow = slow.next
        n += 1
    return p, n

参考答案

-- coding:utf-8 --

class ListNode:

def init(self, x):

self.val = x

self.next = None

class ChkIntersection: def chkInter(self, head1, head2, adjust0, adjust1): # write code here loop1 = self.chkLoop(head1) loop2 = self.chkLoop(head2) if loop1 == loop2: return True cur = loop1.next while cur != loop1: if cur == loop2: return True cur = cur.next return False

def chkLoop(self, head):
    slow, fast = head, head
    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next
        if fast == slow:
            break
    if slow != fast:
        return None
    fast = head
    while fast != slow:
        slow = slow.next
        fast = fast.next
    return fast

单链表相交判断 给定两个单链表的头节点head1和head2，如何判断两个链表是否相交？相交的话返回true，不相交的话返回false。 给定两个链表的头结点head1和head2(注意，另外两个参数adjust0和adjust1用于调整数据,与本题求解无关)。请返回一个bool值代表它们是否相交。 我的提交

-- coding:utf-8 --

class ListNode:

def init(self, x):

self.val = x

self.next = None

class ChkIntersection: def getLoopNode(self, head): fast, slow = head, head while fast and fast.next: fast = fast.next.next slow = slow.next if slow == fast: break if slow != fast: return None fast = head while fast != slow: fast = fast.next slow = slow.next return fast

def chkInter(self, head1, head2, adjust0, adjust1):
    # write code here
    if not head1 or not head2:
        return False
    loop1 = self.getLoopNode(head1)
    loop2 = self.getLoopNode(head2)
    cur1 = head1
    cur2 = head2
    if not loop1 and not loop1:
        while cur1.next:
            cur1 = cur1.next
        while cur2.next:
            cur2 = cur2.next
    if cur1 == cur2:
        return True

    if loop1 == loop2:
        return True
    cur = loop1.next
    while cur != loop1:
        if cur == loop2:
            return True
        cur = cur.next
    return False