PattySung

Given the heads of two singly linked-lists headA and headB, return the node at which the two lists intersect. If the two linked lists have no intersection at all, return null.

For example, the following two linked lists begin to intersect at node c1:

The test cases are generated such that there are no cycles anywhere in the entire linked structure.

Note that the linked lists must retain their original structure after the function returns.

Example 1:

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,6,1,8,4,5], skipA = 2, skipB = 3
Output: Intersected at '8'
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 [5,6,1,8,4,5]. 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 = [1,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
Output: Intersected at '2'
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 [1,9,1,2,4]. 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: No intersection
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.

<解題>

在迴圈中，判斷兩個指針 a 和 b 是否相等。
如果相等，表示找到了交點，跳出迴圈
while( a != b)時，如果短的那條走完，會接著走長的那條，可以確保兩個指針分別走過了鏈表 A 和鏈表 B 的長度之和
若兩者沒有交點，則會在跑完兩個head後，得到值null並返回


public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
    if(headA == null || headB == null) return null;
    
    ListNode a = headA;
    ListNode b = headB;
  
    while( a != b){

        a = a == null? headB : a.next; //如果 a 達到鏈表的尾部，則將 a 設置為 headB，以便後續遍歷 headB 的節點
        b = b == null? headA : b.next; //如果 b 達到鏈表的尾部，則將 b 設置為 headA，以便後續遍歷 headA 的節點
    }
    
    return a;
}

Time : O(m+n) Space : O(1)

Leetcode-160. Intersection of Two Linked Lists

<解題>