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题目：输入员两个递增排序的链表，合并这两个链表并使新的链表中的结点仍然是按照递增排序的。
思路：首先，定义两个头节点分别为Head1和Head2的链表，然后比较第一个节点的值，如果是Head1->mValue比Head2->mValue小，那么头节点，就是Head1，递归实现后面的节点的排序。
C++代码：
#include<iostream> using namespace std; struct ListNode { int m_nValue; ListNode* m_pNext; }; ListNode* CreateList(int a[],int b) { ListNode* pHead=NULL,*pNode=NULL; for(int i=0;i<b;i++) { ListNode* pNew=new ListNode(); pNew->m_nValue=a[i]; pNew->m_pNext=NULL; if(pHead==NULL) { pHead=pNew; pNode=pNew; } else { pNode->m_pNext=pNew; pNode=pNode->m_pNext; } } return pHead; } void PrintList(ListNode* pHead) { if(pHead==NULL) { return; } ListNode* pNode=pHead; while(pNode!=NULL) { cout<<pNode->m_nValue<<" "; pNode=pNode->m_pNext; } cout<<endl; } ListNode* MergeLink(ListNode* Head1,ListNode* Head2) { ListNode* MergeHead=NULL; if(Head1==NULL) return Head2; if(Head2==NULL) return Head1; if(Head1==NULL&&Head2==NULL) return NULL; if(Head1->m_nValue>Head2->m_nValue) { MergeHead=Head2; Head2->m_pNext=MergeLink(Head2->m_pNext,Head1); } else { MergeHead=Head1; Head1->m_pNext=MergeLink(Head1->m_pNext,Head2); } return MergeHead; } void main() { int a[]={3,4,5,6,8}; int b[]={1,2,7}; ListNode* Head1=CreateList(a,5); ListNode* Head2=CreateList(b,3); PrintList(Head1); PrintList(Head2); ListNode* Head3=MergeLink(Head1,Head2); PrintList(Head3); }
Java代码：
public class MergeLink { public static class ListNode { int m_nValue; ListNode m_pNext; }; public static ListNode CreateList(int[] a,int b) { ListNode pHead=null,pNode=null; for(int i=0;i<b;i++) { ListNode pNew=new ListNode(); pNew.m_nValue=a[i]; pNew.m_pNext=null; if(pHead==null) { pHead=pNew; pNode=pNew; } else { pNode.m_pNext=pNew; pNode=pNode.m_pNext; } } return pHead; } public static void PrintList(ListNode pHead) { if(pHead==null) { return; } ListNode pNode=pHead; while(pNode!=null) { System.out.print(pNode.m_nValue+" "); pNode=pNode.m_pNext; } System.out.println(); } public static ListNode MergeLink(ListNode Head1,ListNode Head2) { ListNode MergeHead=null; if(Head1==null) return Head2; if(Head2==null) return Head1; if(Head1==null&&Head2==null) return null; if(Head1.m_nValue>Head2.m_nValue) { MergeHead=Head2; Head2.m_pNext=MergeLink(Head2.m_pNext,Head1); } else { MergeHead=Head1; Head1.m_pNext=MergeLink(Head1.m_pNext,Head2); } return MergeHead; } public static void main(String[] args) { int a[]={3,4,5,6,8}; int b[]={1,2,7}; ListNode Head1=CreateList(a,5); ListNode Head2=CreateList(b,3); PrintList(Head1); PrintList(Head2); ListNode Head3=MergeLink(Head1,Head2); PrintList(Head3); } }
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原文地址：http://www.cnblogs.com/hupp/p/4577278.html
