1/13A.必須是連續的B.局部地址必須是連續的C續的D.連續不連續都可以存儲結構和線性表的鏈式存儲結構分別是〔隨機存取的存儲結02hgbaechfgdbehfca2/13計算機無關的是數據的〔C〕〔B〕3/1324.假設線性表的長度為n，那么在最壞情況下，冒泡排序需要的比擬次數為4/13)。5/13問typedefstructLNode{intdataLNode*next;}LNode,*LinkList;voidReverse(LinkList&head){if(head==NULL)return;while(cur!=NULL){}depqwhile(p!=NULL){}}LNode*Reverse(LNode*p,LinkList&head){if(p==NULL||p->next==NULL){head=p;returnp;6/13}{LNode*tmp=Reverse(p->next,head);tmp->next=p;p->next=NULL;returnp;}}structNode{intdataNode*next;boolIsLoop(Node*head){Node*p1=head,*p2=head;if(head==NULL||head->next==NULL){returnfalse;}{p1=p1->next;p2=p2->next->next;}while(p2&&p2->next&&p1!=p2);if(p1==p2)returntrue;returnfalse;}7/13boolFindCommon(inta[],intsize1,intb[],intsize2){inti;for(i=0;i<size1;i++){intstart=0,end=size2-1,mid;while(start<=end){mid=(start+end)/2;if(a[i]==b[mid])returntrue;elseif(a[i]<b[mid])end=mid-1;start=mid+1;}}returnfalse;}boolFindCommon2(inta[],intsize1,intb[],intsize2){inti=0,j=0;while(i<size1&&j<size2){if(a[i]==b[j])returntrue;if(a[i]>b[j])if(a[i]<b[j])}returnfalse;}8/13環，依次求出所有子序列的和然后取最大的那個。當然算法復雜度會到達O(n^3)。顯然這種方法不是最優的，下面給出一個算法復雜度為O(n)的線性算看一個對窮舉算法進展優化的算法，它的算法復雜度為O(n^2)。其實這個算法算一遍。假設Sum(i,j)是A[i]...A[j]的和，那么Sum(i,j+1)=Sum(i,j)+intMaxSub(inta[],intsize){inti,j,v,max=a[0];for(i=0;i<size;i++){v=0;for(j=i;j<size;j++){v=v+a[j];//Sum(i,j+1)=Sum(i,j)+A[j+1]ifvmax)max=v;}}returnmax;}intMaxSub2(inta[],intsize){inti,max=0,tempSum=0;for(i=0;i<size;i++){tempSum+=a[i];if(tempSum>max)max=tempSum;elseif(tempSum<0)tempSum=0;}returnmax;}9/13串,并不是簡單的字符串反轉，而是按給定字符串里的單詞將H第一個字符和char*ReverseWord(constchar*str){intlen=strlen(str);char*reStr=newchar[len+1];strcpy(reStr,str);inti,j;for(i=0,j=len-1;i<j;i++,j--){chartemp=reStr[i];reStr[i]=reStr[j];reStr[j]=temp;}tkwhile(k<len){while(reStr[j]!=''&&reStr[j]!='\0')k=j+1;j--;for(;i<j;i++,j--){chartemp=reStr[i];reStr[i]=reStr[j];reStr[j]=temp;}}returnreStr;}chartemp=reStr[i];reStr[i]=reStr[j];0/13reStr[j]=temp;reStr[i]^=reStr[j];reStr[j]^=reStr[i];reStr[i]^=reStr[j];字符串的子串，將第一個輸入：第一個字符串:#include<iostream>#include<cassert>#include<stack>usingnamespacestd;//reversethestring's1'exceptthesubstring'token'.constchar*reverse(constchar*s1,constchar*token){assert(s1&&token);stack<char>cStack;constchar*pToken=token,*head=s1,*rear=s1;while(*head!='\0'){while(*head!='\0'&&*pToken==*head){ptoken++;head++;}if(*pToken=='\0')//containthetoken{constchar*p;for(p=head-1;p>=rear;p--)cStack.push(*p);pToken=token;rear=head;}{cStack.push(*rear);head=++rear;pToken=token;}}char*pszRetValue=newchar[strlen(s1)+1];inti=0;while(!cStack.empty()){pszRetValue[i++]=cStack.top();cStack.pop();}pszRetValue[i]='\0';returnpszRetValue;}intmain(intargc,char*argv[]){cout<<"Thisisfishsky'sChinesesite:./cn";coutreverseThisisfishskysChinesesite:./cn","fishsky");return0;}#include<iostream>#include<vector>usingnamespacestd;//removetheduplicatednumbersinanintgerarray,thearraywasendwith0;voidstaticremove_duplicated(inta[],vector<int>&_st){_st.push_back(a[0]);for(inti=1;_st[_st.size()-1]!=0;i++){if(a[i-1]!=a[i])_st.push_back(a[i]);}}2/13voidstaticremove_duplicated2(inta[]){if(a[0]==0||a==NULL)return;intinsert=1,current=1;while(a[current]!=0){if(a[current]!=a[current-1]){a[insert]=a[current];insert;current++;}current++;}a[insert]=0;}的深度相差不template<typenameT>staticintDepth(BSTreeNode<T>*pbs){if(pbs==NULL)return0;{intld=Depth(pbs->left);intrd=Depth(pbs->right);return1+(ld>rd?ld:rd);}}template<typenameT>sta