LeetCode-160. Intersection of Two Linked Lists

LeetCode-160. Intersection of Two Linked Lists

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 = 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 [5,0,1,8,4,5]. 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 = 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 [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 = 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.

题解：

class Solution {public: ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { ListNode *p = headA, *q = headB; int la = 0, lb = 0; while (p != NULL) { p = p->next; la++; } while (q != NULL) { q = q->next; lb++; } if (la < lb) { int dis = lb - la; p = headA, q = headB; for (int i = 0; i < dis; i++) { q = q->next; } while (p != q) { p = p->next; q = q->next; } return p; } else { int dis = la - lb; p = headA, q = headB; for (int i = 0; i < dis; i++) { p = p->next; } while (p != q) { p = p->next; q = q->next; } return p; } return NULL; }};

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