LeetCode-160. Intersection of Two Linked Lists

160. Intersection of Two Linked Lists

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

A:          a1 → a2
                   ↘
                     c1 → c2 → c3
                   ↗            
B:     b1 → b2 → b3

begin to intersect at node c1.

Notes:

If the two linked lists have no intersection at all, return null.

• The linked lists must retain their original structure after the function returns.
• You may assume there are no cycles anywhere in the entire linked structure.
• Your code should preferably run in O(n) time and use only O(1) memory.

Solution

# 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
        """
        # A length = a + c, B length = b + c, after switching pointer, pointer A will move another b + c steps, pointer B will move a + c more steps, 
        # since a + c + b + c = b + c + a + c, it does not matter what value c is. 
        # Pointer A and B must meet after a + c + b (b + c + a) steps. If c == 0, they meet at NULL.
        if not headA or not headB:
            return None
        pA = headA
        pB = headB
        # if the two lists have no intersection at all,you should stop after you've already checked L1+L2, 
        # so we need a flag jumpToNext to ensure we only traverse L1 + L2 once.
        jumpToNext = False
        while pA and pB:
            if pA == pB:
                return pA
            pA, pB = pA.next, pB.next
            if not pA and not jumpToNext:
                pA = headB
                jumpToNext = True
            if not pB:
                pB = headA
        return None

其他方法：

1. 先翻转两个链表，找到汇合节点再将两个链表翻转回来。

2. 先得到两个链表的长度，然后从相同长度的地方开始判断。但是时间复杂度高了。