160.intersection-of-two-linked-lists.0.go

/*
 * @lc app=leetcode id=160 lang=golang
 *
 * [160] Intersection of Two Linked Lists
 *
 * https://leetcode.com/problems/intersection-of-two-linked-lists/description/
 *
 * algorithms
 * Easy (32.92%)
 * Total Accepted:    291.4K
 * Total Submissions: 877.4K
 * Testcase Example:  '8\n[4,1,8,4,5]\n[5,0,1,8,4,5]\n2\n3'
 *
 * 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.
 *
 *
 */
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func getIntersectionNode(headA, headB *ListNode) *ListNode {
	if headA == nil || headB == nil {
		return nil
	}

	nodeA := headA
	nodeB := headB

	count := 0
	overA := false
	lA := 1
	overB := false
	lB := 1

	for count < lA*lB {
		if !overA {
			lA++
		}
		if !overB {
			lB++
		}
		count++

		if nodeA == nodeB {
			return nodeA
		}

		nodeA = nodeA.Next
		nodeB = nodeB.Next

		if nodeA == nil {
			overA = true
			nodeA = headA
		}
		if nodeB == nil {
			overB = true
			nodeB = headB
		}
	}
	return nil
}