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[LeetCode] 160. Intersection of Two Linked Lists（两个单链表的交集）

2020-10-20 09:33:43 阅读：273 来源： 互联网

标签：node intersected linked Two lists two Lists null LeetCode

Difficulty: Easy
Related Topics: Linked List
Link: https://leetcode.com/problems/intersection-of-two-linked-lists/

Description

Write a program to find the node at which the intersection of two singly linked lists begins.

写一个程序，找到两个单链表相交的地方。

For example, the following two linked lists:

例如，以下两个单链表：

begins to intersect at node c1.

在 c1 节点处开始相交。

Examples

Example 1

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,6,1,8,4,5], 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 [5,6,1,8,4,5]. 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: 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 [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: 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.

如果两个单链表毫无交集，返回 null。
The linked lists must retain their original structure after the function returns.

在函数返回后，两个链表必须维持它们原有的结构。
You may assume there are no cycles anywhere in the entire linked structure.

你可以假设在整个结构中不存在环。
Each value on each linked list is in the range [1, 10^9].

链表中每个节点值的范围在 [1, 10^9]。
Your code should preferably run in O(n) time and use only O(1) memory.

你的代码最好有 O(N) 时间复杂度和 O(1) 空间复杂度。

Solution

双指针法。起初我很天真的让 p 和 q 两个指针分别在两个链表上跑，这两指针相同了就表示找到了。但这种做法在两链表不相交时就会进入死循环了，得想其它的办法。那么如果让两个指针跑到尽头时，跳到对方的头节点继续跑会怎样呢？实际试了一下，发现此法不仅可行，且这种“换跑道”的操作刚好只执行一次，遂采取这种方法，代码如下：

/**
 * Example:
 * var li = ListNode(5)
 * var v = li.`val`
 * Definition for singly-linked list.
 * class ListNode(var `val`: Int) {
 *     var next: ListNode? = null
 * }
 */
class Solution {
    fun getIntersectionNode(headA:ListNode?, headB:ListNode?):ListNode? {
        if (headA == null || headB == null) {
            return null
        }

        var p = headA
        var q = headB

        while (p != null && q != null) {
            // 这里用值相等 `==` 实测也能过
            // 不过为了说明“指向同一节点”的语义，这里用引用相等 `===` 更好一点
            if (p === q) {
                return p
            }
            p = p.next
            q = q.next
            // 上面那个“换跑道”的操作，对于不相交的链表，最后 p 和 q 会同时为 null
            if (p == null && q == null) {
                break
            }
            // “换跑道”操作
            if (p == null) {
                p = headB
            }
            if (q == null) {
                q = headA
            }
        }

        return null
    }
}

P.S. 相比较解这道题本身，我觉得从样例给的样例输入里构建这种结构似乎还更难一点，有没有读者想挑战一下？（手动狗头）

标签：node,intersected,linked,Two,lists,two,Lists,null,LeetCode
来源： https://www.cnblogs.com/zhongju/p/13844672.html

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