README_EN.md

160. Intersection of Two Linked Lists

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Description

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:

It is guaranteed 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.

 

Constraints:

• The number of nodes of listA is in the m.
• The number of nodes of listB is in the n.
• 0 <= m, n <= 3 * 104
• 1 <= Node.val <= 105
• 0 <= skipA <= m
• 0 <= skipB <= n
• intersectVal is 0 if listA and listB do not intersect.
• intersectVal == listA[skipA + 1] == listB[skipB + 1] if listA and listB intersect.

 

Follow up: Could you write a solution that runs in O(n) time and use only O(1) memory?

Solutions

Python3

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

class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
        cur1, cur2 = headA, headB
        while cur1 != cur2:
            cur1 = headB if cur1 is None else cur1.next
            cur2 = headA if cur2 is None else cur2.next
        return cur1

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        ListNode cur1 = headA, cur2 = headB;
        while (cur1 != cur2) {
            cur1 = cur1 == null ? headB : cur1.next;
            cur2 = cur2 == null ? headA : cur2.next;
        }
        return cur1;
    }
}

C++

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        ListNode* cur1 = headA;
        ListNode* cur2 = headB;
        while (cur1 != cur2) {
            cur1 = cur1 ? cur1->next : headB;
            cur2 = cur2 ? cur2->next : headA;
        }
        return cur1;
    }
};

JavaScript

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */

/**
 * @param {ListNode} headA
 * @param {ListNode} headB
 * @return {ListNode}
 */
 var getIntersectionNode = function(headA, headB) {
  let cur1 = headA;
  let cur2 = headB;
  while (cur1 != cur2) {
      cur1 = cur1 ? cur1.next : headB;
      cur2 = cur2 ? cur2.next : headA;
  }
  return cur1;
};

Go

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
 func getIntersectionNode(headA, headB *ListNode) *ListNode {
    cur1, cur2 := headA, headB
    for cur1 != cur2 {
        if cur1 == nil {
            cur1 = headB
        } else {
            cur1 = cur1.Next
        }
        if cur2 == nil {
            cur2 = headA
        } else {
            cur2 = cur2.Next
        }
    }
    return cur1
}

TypeScript

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function getIntersectionNode(headA: ListNode | null, headB: ListNode | null): ListNode | null {
    let p1: ListNode | null = headA;
    let p2: ListNode | null = headB;
    while (p1 != p2) {
        p1 = p1 == null ? headB : p1.next;
        p2 = p2 == null ? headA : p2.next;
    }
    return p1;
};

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