[leetcode] 160. Intersection of Two Linked Lists @ python

原题

https://leetcode.com/problems/intersection-of-two-linked-lists/

解法

链表A和链表B的长度可能不相等, 所以我们需要将较长的链表多出的节点忽略掉. 定义getLength函数计算链表的长度, 然后将较长的链表的head移动n次, n为多出的长度. 然后同时遍历链表A和链表B, 当headA == headB时, 返回节点. 这里的edge case是两个链表都只有1个节点且在头部重合, 因此判断重合的条件必须是headA == headB, 而不是headA.next == headB.next
Time: O(m) + O(n) + O(max(m, n)), m, n分别是链表的长度
Space: O(1)

代码

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def getIntersectionNode(self, headA, headB):
        """
        :type head1, head1: ListNode
        :rtype: ListNode
        """
        len_a, len_b = self.getLength(headA), self.getLength(headB)
        if len_a > len_b:
            for i in range(len_a - len_b):
                headA = headA.next
        else:
            for i in range(len_b - len_a):
                headB = headB.next
        
        while headA and headB:
            if headA == headB:
                return headA
            headA = headA.next
            headB = headB.next
        return None      
            
        
    def getLength(self, head):
        l = 0
        while head:
            l += 1
            head = head.next
        return l