[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:

begin to intersect at node c1.

 

Example 1:

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3 Output: Reference of the node with value = 8 Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.

 

Example 2:

Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1 Output: Reference of the node with value = 2 Input Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.

 

Example 3:

Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2 Output: null Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values. Explanation: The two lists do not intersect, so return null.

 

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.

나의 solution :

우선 한번 순회를 통해 A, B의 길이와 각 끝점이 만나는지를 확인한다.

만나지 않으면 둘의 intersection은 없기 때문에 None을 반환한다.

그리고 길이가 긴쪽의 포인터를 차이만큼 움직여서 같은 길이에서 탐색을 시작해서 같은 점을 발견하면 그점이 intersection!

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

class Solution(object):
    
    def changeStart(self,head, cnt):
        while cnt >0:
            head = head.next
            cnt -=1
        return head
        
    def getIntersectionNode(self, headA, headB):
        tailA, tailB = headA, headB
        cntA, cntB = 0, 0
        while tailA :
            tailA = tailA.next
            cntA +=1
        while tailB :
            tailB = tailB.next
            cntB +=1
        if tailA != tailB :
            return None 
        
        if cntA > cntB :
            headA = self.changeStart(headA, cntA-cntB)
        elif cntB > cntA :
            headB = self.changeStart(headB, cntB-cntA)
            
        while headA != headB:
            headA = headA.next
            headB = headB.next
            
        return headA
            
            
        

 

신기한 솔루션

 

class Solution(object):
    def getIntersectionNode(self, headA, headB):
        
        a, b = headA, headB
        while a != b:
            if a is None:
                a = headB
            else:
                a = a.next
            if b is None:
                b = headA
            else:
                b = b.next
        return a