leetcode 160 —— Intersection of Two Linked Lists

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最新推荐文章于 2025-03-18 05:59:11 发布

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本文介绍了一种寻找两个单链表开始相交节点的方法。通过首先计算两个链表的长度，然后让较长链表的头部向前移动两者长度之差的距离，从而确保两个链表剩余部分等长。之后同时遍历两链表直至找到第一个相同节点即为交点。

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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:

A:          a1 → a2
                   ↘
                     c1 → c2 → c3
                   ↗            
B:     b1 → b2 → b3

begin to intersect at node c1.

思路：先计算两个链表的长度，移动较长链表的头指针，使得其后的节点个数与短链表一致

class Solution {
public:
	ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
		int lenA = 0, lenB = 0;
		ListNode *A = headA, *B = headB;
		while (A != nullptr) {
			A = A->next;
			lenA++;
		}
		while (B != nullptr) {
			B = B->next;
			lenB++;
		}
		ListNode *lon, *sho;
		int distance = abs(lenA - lenB);
		if (lenA > lenB) {
			lon = headA;
			sho = headB;
		}
		else {
			lon = headB;
			sho = headA;
		}

		while(distance!=0){
			lon = lon->next;
			distance--;
		}
		while (lon != sho) {
			lon = lon->next;
			sho = sho->next;
		}
		return lon;

	}
};