二つのリンクリストの160交差点
イージー
2つの単独でリンクされたリストの交点が始まるノードを検索するプログラムを書きます。
たとえば、次の2つのリストをリンク:
ノードC1で交差し始めます。
例1:
入力:intersectVal = 8、リスタ= [4,1,8,4,5]、ListBの= [5,0,1,8,4,5]、skipA = 2、SKIPB = 3 出力:値を持つノードのリファレンス= 8 入力説明:交差ノードの値が8である(二つのリストが交差する場合、これは0であってはならないことに注意してください)。Aの頭からは、[4,1,8,4,5]として読み取ります。Bの頭部から、[5,0,1,8,4,5]として読み出します。Aにおける交差ノードの前に2つのノードがあります。B.で交差ノードの前に3つのノードが存在します
例2:
入力:intersectVal = 2、リスタ= [0,9,1,2,4]、ListBの= [3,2,4]、skipA = 3、SKIPB = 1 出力:値を持つノードのリファレンス= 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.
package leetcode.easy; public class IntersectionOfTwoLinkedLists { @org.junit.Test public void test0() { ListNode a1 = new ListNode(1); ListNode a2 = new ListNode(2); ListNode b1 = new ListNode(1); ListNode b2 = new ListNode(2); ListNode b3 = new ListNode(3); ListNode c1 = new ListNode(1); ListNode c2 = new ListNode(2); ListNode c3 = new ListNode(3); a1.next = a2; b1.next = b2; b2.next = b3; a2.next = c1; b3.next = c1; c1.next = c2; c2.next = c3; c3.next = null; System.out.println(getIntersectionNode(a1, b1).val); } @org.junit.Test public void test1() { ListNode a1 = new ListNode(4); ListNode a2 = new ListNode(1); ListNode b1 = new ListNode(5); ListNode b2 = new ListNode(0); ListNode b3 = new ListNode(1); ListNode c1 = new ListNode(8); ListNode c2 = new ListNode(4); ListNode c3 = new ListNode(5); a1.next = a2; b1.next = b2; b2.next = b3; a2.next = c1; b3.next = c1; c1.next = c2; c2.next = c3; c3.next = null; System.out.println(getIntersectionNode(a1, b1).val); } @org.junit.Test public void test2() { ListNode a1 = new ListNode(0); ListNode a2 = new ListNode(9); ListNode a3 = new ListNode(1); ListNode b1 = new ListNode(3); ListNode c1 = new ListNode(2); ListNode c2 = new ListNode(4); a1.next = a2; a2.next = a3; a3.next = c1; b1.next = c1; c1.next = c2; c2.next = null; System.out.println(getIntersectionNode(a1, b1).val); } @org.junit.Test public void test3() { ListNode a1 = new ListNode(2); ListNode a2 = new ListNode(6); ListNode a3 = new ListNode(4); ListNode b1 = new ListNode(1); ListNode b2 = new ListNode(5); a1.next = a2; a2.next = a3; a3.next = null; b1.next = b2; b2.next = null; System.out.println(getIntersectionNode(a1, b1)); } public ListNode getIntersectionNode(ListNode headA, ListNode headB) { if (null == headA || null == headB) { return null; } else { ListNode pA = headA; ListNode pB = headB; while (pA != pB) { if (pA != null) { pA = pA.next; } else { pA = headB; } if (pB != null) { pB = pB.next; } else { pB = headA; } } return pA; } } }