交叉链表

来源：互联网发布：iphone实用软件排行编辑：程序博客网时间：2024/05/16 18:17

两个单向链表，找出它们的第一个公共结点。链表的结点定义为：

struct ListNode

{

int m_nKey;

ListNode* m_pNext;

};

#include "stdafx.h"#include<stack>#include<iostream>using namespace std;struct ListNode{int  m_nKey;ListNode*   m_pNext;};ListNode*find_cross_node(ListNode*head1, ListNode*head2){ListNode*n1, *n2;stack<int>a1, a2;n1 = head1;n2 = head2;while (n1->m_pNext != NULL){a1.push(n1->m_nKey);n1 = n1->m_pNext;}while (n2->m_pNext != NULL){a2.push(n2->m_nKey);n2 = n2->m_pNext;}while (a1.top() == a2.top()){a1.pop();a2.pop();if (a1.empty())break;if (a2.empty())break;}int k;k = a2.size();n2 = head2;while (k){n2 = n2->m_pNext;k--;}k = a1.size();n1 = head1;while (k){n1 = n1->m_pNext;k--;}while (n1 != n2){n1 = n1->m_pNext;n2 = n2->m_pNext;}return n1;}int _tmain(int argc, _TCHAR* argv[]){ListNode*head1 = new ListNode;ListNode*head2 = new ListNode;ListNode*node1 = new ListNode;ListNode*node2 = new ListNode;ListNode*node3 = new ListNode;ListNode*node4 = new ListNode;ListNode*node5 = new ListNode;ListNode*node6 = new ListNode;head1->m_nKey = 0;head2->m_nKey = 0;node1->m_nKey = 0;node2->m_nKey = 0;node3->m_nKey = 0;node4->m_nKey = 4;node5->m_nKey = 5;node6->m_nKey = 6;head1->m_pNext = node1;node1->m_pNext = node2;head2->m_pNext = node3;node3->m_pNext = node4;node2->m_pNext = node4;node4->m_pNext = node5;node5->m_pNext = node6;node6->m_pNext = NULL;ListNode*cross_node = find_cross_node(head1, head2);if (cross_node == NULL)cout << "两个链表不交叉" << endl;system("pause");return 0;}

思路是：从两个链表交叉点开始，后面的都重合，因此很自然将两个链表末尾对齐，对齐方法是从后往前一个节点一个节点的比较值，若不相等，则从这两个链表不相等位置的下一个开始一一比较就行了。

复杂度应该是O(n)。

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