160. Intersection of Two Linked Lists #

题目 #

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.

题目大意 #

找到 2 个链表的交叉点。

解题思路 #

这道题的思路其实类似链表找环。

给定的 2 个链表的长度如果一样长，都从头往后扫即可。如果不一样长，需要先“拼成”一样长。把 B 拼接到 A 后面，把 A 拼接到 B 后面。这样 2 个链表的长度都是 A + B。再依次扫描比较 2 个链表的结点是否相同。

代码 #

package leetcode

import "fmt"

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func getIntersectionNode(headA, headB *ListNode) *ListNode {
	//boundary check
	if headA == nil || headB == nil {
		return nil
	}

	a := headA
	b := headB

	//if a & b have different len, then we will stop the loop after second iteration
	for a != b {
		//for the end of first iteration, we just reset the pointer to the head of another linkedlist
		if a == nil {
			a = headB
		} else {
			a = a.Next
		}

		if b == nil {
			b = headA
		} else {
			b = b.Next
		}
		fmt.Printf("a = %v b = %v\n", a, b)
	}
	return a
}