zoukankan      html  css  js  c++  java
  • 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:

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

    begin to intersect at node c1.

    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.
    • Your code should preferably run in O(n) time and use only O(1) memory.

     

    Credits:
    Special thanks to @stellari for adding this problem and creating all test cases.

    这道题是说寻找两个链表的公共部分的起始节点,有这么几个方法:

    1、暴力搜索

    时间复杂度太高

    2、使用hash表

    空间复杂度不满足要求

    3、双指针

    方法非常巧妙,设置两个指针分别从两个链表头部开始遍历,如果某个指针到达尾部,则从另一个链表头部开始,最后两个指针相遇的地方就是公共节点

     1 /**
     2  * Definition for singly-linked list.
     3  * struct ListNode {
     4  *     int val;
     5  *     ListNode *next;
     6  *     ListNode(int x) : val(x), next(NULL) {}
     7  * };
     8  */
     9 class Solution {
    10 public:
    11     ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
    12         if (!headA || !headB)
    13             return nullptr;
    14         ListNode *a = headA, *b = headB;
    15         while (a != b)
    16         {
    17             a = a ? a->next : headB;
    18             b = b ? b->next : headA;
    19         }
    20         return a;
    21     }
    22 };

    时间复杂度:O(m+n)

    空间复杂度:O(1)

    参考:https://leetcode.com/problems/intersection-of-two-linked-lists/solution/

    https://leetcode.com/problems/intersection-of-two-linked-lists/discuss/49799

  • 相关阅读:
    【程序2】
    【程序1】
    基数排序
    归并排序
    选择排序
    leetcode15
    leetcode221
    leetcode322
    leetcode921
    leetcode922
  • 原文地址:https://www.cnblogs.com/dapeng-bupt/p/8276397.html
Copyright © 2011-2022 走看看