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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 = [5,6,1,8,4,5], skipA = 2, skipB = 3Output: Reference of the node with value = 8Input 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 = 1Output: Reference of the node with value = 2Input 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 = 2Output: nullInput 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.
• Each value on each linked list is in the range [1, 10^9].
• Your code should preferably run in O(n) time and use only O(1) memory.

题意：编写一个满足 O(n) 时间复杂度、且仅用 O(1) 内存的程序，找到两个单链表相交的起始节点（注意，整个链表结构中没有环）。如果两个链表没有交点，返回 null 。在返回结果后，两个链表仍须保持原有的结构。

解法 双指针

这里我们定义两个指针。第一轮让两个指针不断前进，到达末尾节点后使其指向另一个链表的头部。接着是第二轮移动，如果相遇则为交点，有交点就返回；无交点则最终两个指针都等于 nullptr ，它们相等从而退出循环。

class Solution {
   public:    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
           if (headA == nullptr || headB == nullptr) return nullptr; //无交点        ListNode *pa = headA, *pb = headB;         while (pa != pb) {
               pa = (pa == nullptr) ? headB : pa->next;            pb = (pb == nullptr) ? headA : pb->next;        }        return pa;    }};

为什么这样做是正确的？因为两个指针都完成第一轮移动后，恰好抹除了长度差。以下图为例，第一轮移动后，pb 先完成第一轮移动并指向 0 ，此时 pa 指向 2 ；当 pa 完成第一轮移动并指向 3 时，pb 指向 1 。显然，它们再次移动，一定会相遇。

执行用时：48 ms, 在所有 C++ 提交中击败了94.76% 的用户内存消耗：14.7 MB, 在所有 C++ 提交中击败了16.03% 的用户

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