相交链表判断

标签：while   next   指针   ret   ict   one   dict   链表   else   

方法1：

哈希表

时间复杂度：O(m+n)

空间复杂度：O(m)或O(n)

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
        dict =  {}
        cur1 = headA
        while cur1:
             dict[cur1] = 1
             cur1 = cur1.next
        cur2 = headB
        while cur2:
            if cur2 in dict:
                return cur2
            else:
                dict[cur2] = 1
            cur2 = cur2.next
        return None

方法2：

双指针

时间复杂度：O(m+n)

空间复杂度：O(1)

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
        cur1 = headA
        cur2 = headB
        while cur1!=cur2:
            cur1 = cur1.next if cur1 else headB
            cur2 = cur2.next if cur2 else headA
        return cur1

相交链表判断

标签：while   next   指针   ret   ict   one   dict   链表   else   

原文地址：https://www.cnblogs.com/gugu-da/p/13191679.html