leetcode 160. 相交链表(Intersection of Two Linked Lists)

目录

  • 题目描述:
  • 示例 1:
  • 示例 2:
  • 示例 3:
  • 注意:
  • 解法:

题目描述:

编写一个程序,找到两个单链表相交的起始节点。

如下面的两个链表:

在节点 c1 开始相交。

示例 1:

    输入:intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
    输出:Reference of the node with value = 8
    输入解释:相交节点的值为 8 (注意,如果两个列表相交则不能为 0)。
        从各自的表头开始算起,链表 A 为 [4,1,8,4,5],链表 B 为 [5,0,1,8,4,5]。
        在 A 中,相交节点前有 2 个节点;在 B 中,相交节点前有 3 个节点。

示例 2:

    输入:intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
    输出:Reference of the node with value = 2
    输入解释:相交节点的值为 2 (注意,如果两个列表相交则不能为 0)。
        从各自的表头开始算起,链表 A 为 [0,9,1,2,4],链表 B 为 [3,2,4]。
        在 A 中,相交节点前有 3 个节点;在 B 中,相交节点前有 1 个节点。

示例 3:

    输入:intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
    输出:null
    输入解释:从各自的表头开始算起,链表 A 为 [2,6,4],链表 B 为 [1,5]。
        由于这两个链表不相交,所以 intersectVal 必须为 0,而 skipA 和 skipB 可以是任意值。
    解释:这两个链表不相交,因此返回 null。

注意:

  • 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。
  • 可假定整个链表结构中没有循环。
  • 程序尽量满足 O(n) 时间复杂度,且仅用 O(1) 内存。

解法:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        ListNode * h1 = headA;
        ListNode * h2 = headB;
        if(!h1 || !h2){
            return NULL;    // there has an empty linkedlist
        }
        while(h1 && h2 && h1 != h2){
            h1 = h1->next;
            h2 = h2->next;
            if(h1 == h2){
                break;
            }
            if(!h1){
                h1 = headB;
            }
            if(!h2){
                h2 = headA;
            }
        }
        return h1;  // if h1 != NULL, indicates the two nodes intersect
    }
};

原文地址:https://www.cnblogs.com/zhanzq/p/10563156.html

时间: 2024-11-02 12:32:54

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