面试OR笔试21——两个链表的第一个公共节点

1 题目及要求

1.1 题目描述

输入两个链表，找出他们的第一个公共节点。

 

2 解答

2.1 代码

#include <iostream>#include <stack>using namespace std;// 链表节点类struct ListNode{int val;ListNode *next;ListNode(int x = 0):val(x),next(nullptr){}};// 扫描 o(m*n)时间ListNode* firstCommonNode1(ListNode *head1, ListNode *head2){if(!(head1&&head2)) return nullptr;for(ListNode* np;head1;head1 = head1->next)for(np = head2;np;np = np->next)if(np==head1) return np;return nullptr;}// 栈实现,从尾到头逐个比较,直到最后一个相同的 o(m+n)时间, o(m+n)空间ListNode* firstCommonNode2(ListNode *head1, ListNode *head2){if(!(head1&&head2)) return nullptr;stack<ListNode*> st1,st2;for(;head1;head1 = head1->next) st1.push(head1);for(;head2;head2 = head2->next) st2.push(head2);while(!(st1.empty() || st2.empty()) && st1.top()==st2.top()){head1 = st1.top();st1.pop();st2.pop();}return head1;}// 循环实现, o(m+n)时间, o(1)空间ListNode* firstCommonNode3(ListNode *head1, ListNode *head2){if(!(head1&&head2)) return nullptr;ListNode *np1 = head1, *np2 = head2;while(np1!=np2){np1 = np1 ? np1->next : head2;np2 = np2 ? np2->next : head1;}return np1;}int main(){ListNode h1[] = {1,2,3,4,5,6,7}, h2[] = {8,9};int s1 = sizeof(h1)/sizeof(ListNode), s2 = sizeof(h2)/sizeof(ListNode);for(int k1(0);k1<s1-1;++k1) h1[k1].next = h1+k1+1;for(int k2(0);k2<s2-1;++k2) h2[k2].next = h2+k2+1;// 这里从第5个节点(val = 5)开始公用节点h2[s2-1].next = h1+4;auto np1 = firstCommonNode1(h1,h2);auto np2 = firstCommonNode2(h1,h2);auto np3 = firstCommonNode3(h1,h2);cout << (np1?np1->val:0)  << endl;cout << (np2?np2->val:0)  << endl;cout << (np3?np3->val:0)  << endl;    return 0;}