0160. Intersection of Two Linked Lists (E)

Intersection of Two Linked Lists (E)

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.

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题意

求两个链表的交点。

思路

最简单的方法hash：先遍历一遍A链表，将所有结点存入set，再遍历一遍B链表，判断当前结点是否在set中。

也可以通过计算长度来解题：先遍历得到链表A和B的长度，计算出长度差diff，在较长链表中将指针移至第diff个结点处，然后同时移动两链表的指针，当第一次指向同一个结点时就是所要求的的交点。

官方还提供了Two Pointers方法：指针pA和pB分别指向链表A和B的头结点，同时开始向后移动；如果pA遍历完了当前链表，则将其重新指向另一个链表的头结点，对pB同样如此；任何时候只要pA和pB指向同一个结点，就是所求的交点(因为在相遇时，两指针走过的路程长度是一样的)。

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代码实现

Java

hash

public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        Set<ListNode> set = new HashSet<>();
        while (headA != null) {
            set.add(headA);
            headA = headA.next;
        }
        while (headB != null) {
            if (set.contains(headB)) {
                return headB;
            }
            headB = headB.next;
        }

        return null;
    }
}

计算长度

public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        int lenA = 0, lenB = 0;
        ListNode p = headA, q = headB;
        
        while (p != null) {
            lenA++;
            p = p.next;
        }
        while (q != null) {
            lenB++;
            q = q.next;
        }
        
        int diff = Math.abs(lenA - lenB);
        while (diff > 0) {
            if (lenA > lenB) {
                headA = headA.next;
            } else {
                headB = headB.next;
            }
            diff--;
        }
        
        while (headA != headB) {
            headA = headA.next;
            headB = headB.next;
        }
        
        return headA;
    }
}

Two Pointers

public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        ListNode pA = headA, pB = headB;
        while (pA != pB) {
            pA = pA == null ? headB : pA.next;
            pB = pB == null ? headA : pB.next;
        }
        return pA;
    }
}

JavaScript

/**
 * @param {ListNode} headA
 * @param {ListNode} headB
 * @return {ListNode}
 */
var getIntersectionNode = function (headA, headB) {
  const set = new Set()
  while (headA) {
    set.add(headA)
    headA = headA.next
  }
  while (headB) {
    if (set.has(headB)) return headB
    headB = headB.next
  }
  return null
}