剑指offer-面试题37.两个链表的第一个公共结点

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题目:输入两个链表，找出它们的第一个公共结点，链表结点的定义如下:

1 struct ListNode2 {3      int     m_nKey;4      ListNode* m_pNext;  5 }

一共三种方法解答此题目:

方法1:双层遍历链表,复杂度为O(m*n)

方法2:有两个栈存储链表结点模拟从后往前遍历

方法3:先计算出两个链表长度，然后使得较长的链表先走k步与短链表对齐再一起向后比较

这里我们只介绍方法三

我们以链表:

1->2->3->6->7

4->5->6->7为例

1.先计算出链表1长度为5链表2长度为4

2.链表1比链表2长一个结点那么让链表1先走一步到结点2

3.两个链表同时向后遍历并比较遇到相同的结点返回并结束

简单实现如下:

  1 #include <iostream>  2 using namespace std;  3   4 struct ListNode  5 {  6      int     m_nKey;  7      ListNode* m_pNext;    8 };  9  10 void PrintListNode(ListNode* Head) 11 { 12     ListNode* Temp=Head; 13     while(Temp!=NULL) 14     { 15         cout<<Temp->m_nKey<<","; 16         Temp=Temp->m_pNext; 17     } 18     cout<<endl; 19 } 20  21  22 ListNode* FindTheFirstSameNode(ListNode*Head1,ListNode* Head2) 23 { 24     if(Head1==NULL||Head2==NULL) 25         return NULL; 26  27     ListNode* TempHead1=Head1; 28     ListNode* TempHead2=Head2; 29      30     int len1=0; 31     int len2=0; 32  33     while(TempHead1!=NULL) 34     { 35         len1++; 36         TempHead1=TempHead1->m_pNext; 37     } 38     while(TempHead2!=NULL) 39     { 40         len2++; 41         TempHead2=TempHead2->m_pNext; 42     } 43  44     int k=0; 45     TempHead1=Head1; 46     TempHead2=Head2; 47     if(len1>len2) 48     { 49         k=len1-len2; 50         TempHead1=Head1; 51         int count=0; 52         while(count<k) 53         { 54             TempHead1=TempHead1->m_pNext; 55             count++; 56         } 57     } 58     if(len2>len1) 59     { 60         k=len2-len1; 61         TempHead2=Head2; 62         int count=0; 63         while(count<k) 64         { 65             TempHead2=TempHead2->m_pNext; 66             count++; 67         } 68     } 69  70  71     while(TempHead1!=NULL&&TempHead2!=NULL) 72     { 73         if(TempHead1->m_nKey==TempHead2->m_nKey) 74              return TempHead1; 75          76         TempHead1=TempHead1->m_pNext; 77         TempHead2=TempHead2->m_pNext; 78     } 79  80     return NULL; 81 } 82  83  84 int main() 85 { 86     ListNode *Head1; 87     Head1=new ListNode(); 88     Head1->m_nKey=1; 89     Head1->m_pNext=NULL; 90  91     ListNode *Node1=new ListNode(); 92     ListNode *Node2=new ListNode(); 93     ListNode *Node3=new ListNode(); 94     ListNode *Node4=new ListNode(); 95  96     Node1->m_nKey=2; 97     Node1->m_pNext=NULL; 98      99 100     Node2->m_nKey=3;101     Node2->m_pNext=NULL;102 103     Node3->m_nKey=6;104     Node3->m_pNext=NULL;105 106     Node4->m_nKey=7;107     Node4->m_pNext=NULL;108 109     Head1->m_pNext=Node1;110     Node1->m_pNext=Node2;111     Node2->m_pNext=Node3;112     Node3->m_pNext=Node4;113     114     ListNode *Head2=new ListNode();115     Head2->m_nKey=4;116     Head2->m_pNext=NULL;117     118     ListNode *Node5=new ListNode();119     Node5->m_nKey=5;120     Node5->m_pNext=NULL;121 122     Head2->m_pNext=Node5;123 124 125     Node5->m_pNext=Node3;126     cout<<"PrintListNode One: ";127     PrintListNode(Head1);    128 129     cout<<"PrintListNode Two: ";130     PrintListNode(Head2);    131 132     ListNode* FirstNode;133     FirstNode=FindTheFirstSameNode(Head1,Head2);134 135     cout<<"The First SameNode is: "<<FirstNode->m_nKey<<endl;136     return 0;137 }

运行截图:

代码中结点的释放读者自行释放即可。

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