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每日微软面试题
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每日微軟面試題——day 1 <以下微軟面試題全來自網絡>
<以下答案與分析純屬個人觀點，不足之處，還望不吝指出^_^>
題：.編寫反轉字符串的程序，要求優化速度、優化空間。
分析
實現代碼：
[cpp] view plaincopy
? ? ? ?? ?? #include?<stdio.h> ??void ?reverse(char ?*_str,int ?_l)???{?? ?char ?*p=_str,*q=_str+_l-1,temp;?? ?? ?while (q>p)?? ???{??? ?????temp=*p;?? ?????*p=*q;?? ?????*q=temp;?? ??? ?????p++;?? ?????q--;?? ???}?? }?? ??? int ?main()??{?? ??charstr0[11]=?"0123456789" ;??????? ?reverse(str0,sizeof (str0)-1);?? ?printf("str0?=?%s\n" ,str0);?? ??? ??char ?str1[6]="01234" ;?? ?reverse(str1,sizeof (str1)-1);?? ?printf("str1?=?%s" ,str1);?? ??return ?0;?? }??
題：.在鏈表里如何發現循環鏈接？
分析：
實現代碼：
[cpp] view plaincopy
? ? ? ?? #include?<stdio.h> ??#include?<malloc.h> ???? #define?TEST ???? struct ?list_node??{?? ??int ?data;?? ??list_node?*?next;?? };?? list_node?*head;??? ?? void ?list_create()??{?? ??int ?i;?? ??list_node?*p=NULL;?? ??head?=?(list_node*)malloc(sizeof (list_node));?? ??head->data=0;??????? ??head->next?=?NULL;?? ?? ??p=head;?? ??for (i=1;?i<6;?i++)??? ????{?? ??????p->next?=?(list_node*)malloc(sizeof (list_node));?? ??????p->next->data?=?i;?? ??????p->next->next?=?NULL;?? ??????p=p->next;?? ????}?? ??p->next?=?head;???? ?? #ifdef?TEST ????p=head;?? ??for (i=0;?i<12&&p!=NULL;?i++)?? ????{?? ??????printf("%d?" ,p->data);?? ??????p=p->next;?? ????}?? ??printf("\n" );?? #endif ??}?? ?? void ?cycleList_test()??{?? ??if (head==NULL?||?head->next?==?NULL)?? ??????return ;?? ??list_node?*p=NULL,*pp=NULL;?? ??p=head;?? ??pp=head->next->next;?? ??while (p!=pp?)?? ????{?? ??????p=p->next;?? ??????pp=pp->next->next;?? ?? ??????if (p==NULL?||?pp==NULL)?? ????{?? ??????printf("不是循環鏈表" );?? ??????return ?;?? ????}?? ????}?? ??printf("是循環鏈表" );?? }?? ?? void ?list_destroy()??{?? ??list_node?*pmove=NULL,*pdel=NULL;?? ??pmove=head->next;?? ?? ?while (pmove!=head)?? ???{?? ?????pdel=pmove;?? ?????pmove=pmove->next;?? ?????free(pdel);????? ???}?? ?? ??free(head);?? }?? int ?main()??{?? ??list_create();????? ??cycleList_test();?? ??list_destroy();???? ??return ?0;?? }? 題：.給出洗牌的一個算法，并將洗好的牌存儲在一個整形數組里。
分析： 首先54張牌分別用0到53 的數值表示并存儲在一個整形數組里，數組下標代表紙牌所在的位置。接下來，遍歷整個數組，在遍歷過程中隨機產生一個隨機數，并以該隨機數為下標的數組元素與當前遍歷到的數組元素進行對換。時間復雜度為O(n) (注：所得到的每一種結果的概率的分母越大越好)
實現代碼：
[cpp] view plaincopy
? ? ? ?? ?? #include?<stdio.h> ??#include?<time.h> ??#include?<stdlib.h> ???? void ?shuffle(int ?boke[])????{?? ??int ?i,r,t;?? ??srand((unsigned)time(NULL));??? ??for (i=0;?i<54;?i++)?? ????{?? ??????r=(rand()%107)/2;?? ?? ???????? ??????t=boke[i];?? ??????boke[i]=boke[r];?? ??????boke[r]=t;?? ????}?? }?? ?? int ?main()??{?? ??int ?boke[54],i;?? ??for (i=0;i<54;i++)??? ????boke[i]=i;?? ?? ??printf("before?shuffle:\n" );?? ??for (i=0;?i<54;?i++)?????? ??????printf("%d?" ,boke[i]);?? ?? ?? ??shuffle(boke);??????? ?? ?? ??printf("\nafter?shuffle:\n" );?? ??for (i=0;?i<54;?i++)????? ??????printf("%d?" ,boke[i]);?? ?? ??return ?0;?? }?? 題：寫一個函數，檢查字符串是否是整數，如果是，返回其整數值。 （或者：怎樣只用4行代碼編寫出一個從字符串到長整形的函數？）
分析 ：略
實現代碼：
[cpp] view plaincopy
? ? ? ?? ?? #include?<stdio.h> ???? long ?convert(const ?char ?*str)???{?? ??long ?value=0,f=1;??????? ?? ??if (*str?==?'-' )?str++,f=-1;?? ?? ??for (;*str!='\0' ?&&?*str>='0' ?&&?*str<='9' ;?str++)?? ????value=value*10+(*str-'0' );???? ?? ??return ?*str=='\0' ?value*f:0;???? }?? ?? int ?main()??{?? ?? ??char ?str0[]?=?"1234567" ;?? ??printf("%d\n" ,convert(str0));?? ?? ??char ?str1[]?=?"4.56" ;?? ??printf("%d\n" ,convert(str1));?? ?? ??char ?str2[]?=?"-1234567" ;?? ??printf("%d\n" ,convert(str2));?? ?? ??char ?str3[]?=?"-4.56" ;?? ??printf("%d\n" ,convert(str3));?? ???? ??return ?0;?? }? 題：怎樣從頂部開始逐層打印二叉樹結點數據？請編程。
分析 ：不用遞歸，定義兩個棧(或數組)，也能方便漂亮地搞定。首先棧1存儲第一層中的節點也就是根節點，然后每次循環，打印棧1中的元素，再將棧1中的節點更新為下一層中的節點。總共循環logn+1次。
實現代碼（以下遺漏了對二叉樹的銷毀操作）：
[cpp] view plaincopy
? ? ? ?? #include?<stdio.h> ??#include?<malloc.h> ???? typedef ?struct ?bt_node??{?? ??int ?data;?? ??struct ?bt_node?*rchild;?? ??struct ?bt_node?*lchild;?? }BinTree,node_t;?? ?? BinTree?*myTree;?? ?? node_t*?bt_new_node(int ?data)?? {?? ??node_t*?node?=?(node_t*)malloc(sizeof (node_t));?? ??node->rchild?=?NULL;?? ??node->lchild?=?NULL;?? ??node->data?=?data;?? ?? ??return ?node;?? }?? void ?bt_create()??{?? ???? ??myTree?=?bt_new_node(11);?? ?? ???? ??myTree->lchild?=?bt_new_node(21);?? ??myTree->rchild?=?bt_new_node(22);??? ?? ?? ???? ??myTree->lchild->lchild?=?bt_new_node(31);?? ??myTree->lchild->rchild?=?bt_new_node(32);?? ??myTree->rchild->lchild?=?bt_new_node(33);?? ??myTree->rchild->rchild?=?bt_new_node(34);?? ?? ???? ??myTree->rchild->rchild->rchild?=?bt_new_node(48);?? }?? ?? void ?print_layer_by_layer()????{?? ??node_t*?stack1[100]={0};?? ??node_t*?stack2[100]={0};?? ??int ?T1=0,T2=0;????????????? ?? ?? ??stack1[T1++]=myTree;????? ??while (T1)??? ????{?? ??????while (T1)?? ????{?? ??????T1--;?? ??????printf("%d?" ,stack1[T1]->data);??? ??????stack2[T2++]=stack1[T1];?????????? ????}?? ??????printf("\n" );?? ?? ???????? ???????? ??????while (T2)???? ????{?? ??????T2--;?? ??????if (stack2[T2]->rchild?!=?NULL)?? ????????stack1[T1++]=stack2[T2]->rchild;?? ?? ??????if (stack2[T2]->lchild?!=?NULL)?? ????????stack1[T1++]=stack2[T2]->lchild;?? ????}?? ????}?? }?? int ?main()??{?? ??bt_create();?????????????? ??print_layer_by_layer();??? ??return ?0;?? }?? 另： 關于此題的其它想法請看3樓我的回復，回復上的編碼步驟為廣度優先遍歷二叉樹算法， 不過本人不才哈，不知題目中的意思是一層一層的打印呢，還是按照層的順序從左到右一個一個打印呢（好像也是逐層打印），且不管它，通過對兩種算法比較，上面使用棧的算法功能更強，至少打完一層可以再打一個回行。而使用隊列的廣度優先遍歷二叉樹算法 有效率優勢，且代碼簡潔易懂，就是不能打完一層后回行。^_^ 純屬個人觀點，望不吝指教。
題：怎樣把一個鏈表掉個順序（也就是反序，注意鏈表的邊界條件并考慮空鏈表）？
分析： 這題比較有意思，我想了個高效的實現。首先定義兩個迭代器 p 和 q，q從第一個節點開始遍歷，p從第二個節點開始遍歷，每次遍歷將頭指針的next指向p的next，然后將p的next 反指向q(此時q是p的前一個節點)，也就是說將每個節點的鏈接方向掉過來，最后尾節點變成了頭節點，頭節點變成了尾節點，時間復雜度為高效的O(n)
圖示：(最后一個N指空節點)
以下便是是我簡潔的實現代碼：
[cpp] view plaincopy
? ? ? ?? ??? #include?<stdio.h> ??#include?<malloc.h> ????? #define?TEST ????? typedef ?struct ?list_node??{?? ??int ?data;?? ??structlist_node?*?next;?? }list_node;?? ??? list_node?*head;??? ??? void ?list_create()??{?? ??int ?i;?? ??list_node?*p;?? ??head?=?(list_node*)malloc(sizeof (list_node));?? ??head->data=0;??????? ??head->next?=?NULL;?? ??? ??p=head;?? ??for (i=1;i<6;?i++)??? ????{?? ??????p->next?=?(list_node*)malloc(sizeof (list_node));?? ??????p->next->data?=?i;?? ??????p->next->next?=?NULL;?? ??????p=p->next;?? ????}?? ??? #ifdef?TEST ????p=head;?? ??while (p!=NULL)????? ????{?? ??????printf("%d" ,p->data);?? ??????p=p->next;?? ????}?? ??printf("\n" );?? #endif ??}?? ??? void ?list_reverse()????{?? ??printf("^^after?reversing:\n" );?? ??if (head?==?NULL?||?head->next==NULL)?return ?;?? ??list_node?*p,*q;?? ??q=head;?? ??p=head->next;?? ??? ??while (p!=NULL)?? ????{?? ????head->next=p->next;??? ????p->next=q;???????????? ????q=p;?????????????????? ????p=head->next;????????? ????}?? ????? ????head?=?q;?? ??? #ifdef?TEST ????p=head;?? ??while (p!=NULL)?????????? ????{?? ??????printf("%d" ,p->data);?? ??????p=p->next;?? ????}?? ??printf("\n" );?? #endif ????? }?? ??? void ?list_destroy()????{?? ??list_node?*pmove=NULL,*pdel=NULL;?? ??pmove=head;?? ??? ?while (pmove!=head)?? ???{?? ?????pdel=pmove;?? ?????free(pdel);???? ?????pmove=pmove->next;?? ???}?? }?? int ?main()??{?? ??list_create();????? ??list_reverse();???? ??list_destroy();???? ??return ?0;?? }? 題： 求隨機數構成的數組中找到長度大于=3的最長的等差數列
輸出等差數列由小到大:?
如果沒有符合條件的就輸出[0,0]
格式：
輸入[1,3,0,5,-1,6]
輸出[-1,1,3,5]
要求時間復雜度，空間復雜度盡量小
分析 ：基本算法思路（采用動態規劃思想）：首先，只要得到數列的公差和一個首項就可以確定一個等差數列 ，因此我們要尋找最長等差數列的公差以及首項。其次，為了方便查找公差和首項，我們應該將原數組進行由小到大排序 ，這樣各兩數之間的公差也是成遞增形勢的，這樣我們就可以避免回溯查找首項 。
因此，在搜尋公差及首項的過程中，我們可以分兩三個決策階段：
1、如果公差為0 ，應該做何處理。
2、如果公差不為0 ，應該做何處理。
3、如果找到的數列長度是當前最長的 做相應的處理
? ? ? 針對以上情況，我們應該選擇一種合適的數據結構——平衡排序樹，stl中的set，map，mutiset，multimap是以紅黑樹結構為形勢的容器 ，無疑是非常合適的，根據題目情況，可能存在具有相同元素的數組，因此我們選擇multiset，這樣無論我們對數據進行插入排序，查找都是比較高效的， 因此總體上是可以滿意的。
? ? ? 最后有幾項時間上的優化不在這里說明，詳情可看代碼。若有不足之處，望能不吝指出！^_^
我的實現代碼：
[cpp] view plaincopy
? ? ? ?? #include?<iostream> ??#include?<ctime> ??#include?<set> ??using ?namespace ?std;???? void ?show_longest_seq(const ?multiset<int >&?myset)??{?? ????int ?maxLength?=?0,?curr_pos?=?0,?curr_d?=?0,?counter=0,i=0;??? ????int ?d_result,?a1_result;??? ????multiset<int >::const_iterator?set_it1,set_it2;?? ?? ?????? ????? ? ?? ????for (set_it1?=?myset.begin();?set_it1?!=?myset.end();)?? ????{?? ????????for (set_it2=set_it1,set_it2++;?set_it2?!=?myset.end();)?? ????????{?? ????????????curr_d?=?*set_it2?-?*set_it1;??? ?? ????????????if (curr_d?==?0)??? ????????????{?? ????????????????counter?=?myset.count(*set_it1);?? ????????????????set_it2?=?myset.upper_bound(*set_it1);?? ????????????}?? ????????????else ?? ????????????{?? ????????????????counter?=?2;??? ????????????????while (myset.find(*set_it1?+?counter*curr_d)?!=?myset.end())??? ????????????????????++counter;?? ?? ????????????????set_it2?=?myset.upper_bound(*set_it2);?? ????????????}?? ?? ?????????????? ????????????if (counter?>?maxLength)???? ????????????{?? ????????????????d_result?=?curr_d;?? ????????????????a1_result?=?*set_it1;?? ????????????????????????maxLength?=?counter;?? ????????????}?? ????????}?? ?? ????????curr_pos?+=?myset.count(*set_it1);??????? ????????if (myset.size()-curr_pos?<?maxLength)???? ????????????break ;?? ?? ????????set_it1?=?myset.upper_bound(*set_it1);??? ????}?? ?? ?? ?? ?? ????? ? ?? ????if (maxLength?<=?2)?? ????{?? ????????cout<<"longest_seq：[0,0]" <<endl;?? ????}?? ????else ?? ????{?? ????????cout<<"longest_seq：" ;?? ?????????? ????????for (i?=?0;?i<maxLength;??i++)?? ????????????cout<<*(myset.find(a1_result?+?i*d_result))<<'?' ;?? ?? ????????cout<<endl;?? ????}?? }?? ?? ?? int ?main()??{?? ????int ?a[]={1,3,0,5,-1,6};?? ????multiset<int >?myset;?? ????myset.insert(a,a+6);?? ????show_longest_seq(myset);?? ????cout<<endl;?? ?? ????int ?l;?? ????srand((unsigned)time(NULL));?? ????for (int ?j?=?0;?j?<?5;?j++)?? ????{?? ????????myset.clear();?? ????????cout<<"input:[?" ;?? ????????l=rand()%10;?? ????????for (int ?i?=?0;?i?<?l;?++i)?? ????????{?? ????????????int ?element?=?rand()%10;?? ????????????myset.insert(element);?? ????????????cout<<element<<'?' ;?? ????????}?? ????????cout<<']' <<endl;?? ????????show_longest_seq(myset);?? ????????cout<<endl;?? ????}?? ?? ?? ????return ?0;?? }?? 附一張測試結果圖：
題： 兩個鏈表，一升一降。合并為一個升序鏈表。
分析 ： （ 假設升序的鏈表為鏈表1，降序的鏈表為鏈表2，p1，p2分別作為它們的迭代器，還有一個合并鏈表用于存放合并后的數據 ）
法一 、最容易想到的且容易實現的就是使兩個表都變成升序，然后就是經典的合并排序算法的步驟了 ，步驟是構建p1，p2兩個迭代器，分別用于迭代兩個鏈表，每次遍歷，若p1所指的節點數據大于p2所指的節點數據則將p1所指的節點數據插入到要合并鏈表中去且使p1指向下一個節點，否則將p2將所指的節點數據插入到合并鏈表中去且使p2指向下一個節點，直到p1和p2任意一個指向了NULL為止。最后可能兩個鏈表中尚有剩余的節點，將其逐個插入到合并鏈表中去即可。
法二 、使用遞歸方法后序遍歷降序鏈表2，遍歷順序就相當于升序的順序了。在遞歸遍歷鏈表2的過程中，需要處理有以下三件事：(注意順序)
(1) 如果p2的數據小于p1的就插入到合并鏈表中 (2) 如果p2的數據大于p1，那么就對鏈表1循環遍歷，每次將p1中的數據插到合并鏈表中，直到p2不大于p1，且p1不為空 (3) 如果p1為空，就直接將p2插入到合并鏈表中
(這個方法你想到了沒！)
完整實現代碼： [cpp] view plaincopy
? ? ? ?? ??? #include?<stdio.h> ??#include?<malloc.h> ??#include?<stdlib.h>? ???? typedef ?struct ?list_node??{?? ??int ?data;?? ??struct ?list_node?*?next;?? }list_node;?? ??? list_node?*list1=NULL;??? list_node?*list2=NULL;??? ?? void ?list_print(const ?list_node?*p)??{?? ??if (p==NULL)return ;?? ??while (p!=NULL)????? ????{?? ??????printf("%d?" ,p->data);?? ??????p=p->next;?? ????}?? ??printf("\n" );?? }?? ?? void ?list_create(list_node*?&head,?int ?data[],?int ?N)??{?? ??if (data?==?NULL)?return ;?? ?? ??int ?i;?? ??list_node?*p;?? ??p?=?(list_node*)malloc(sizeof (list_node));?? ??p->data?=?data[0];?? ??p->next?=?NULL;?? ??head?=?p;??? ??for (i=1;i<N;?i++)??? ????{?? ??????p->next?=?(list_node*)malloc(sizeof (list_node));?? ??????p->next->data?=?data[i];?? ??????p->next->next?=?NULL;?? ??????p=p->next;?? ????}?? }?? ?? void ?list_reverse(list_node*?&head)????{?? ??if (head?==?NULL)?return ?;?? ??list_node?*p,*q;?? ??q=head;?? ??p=head->next;?? ??? ??while (p!=NULL)?? ????{?? ????head->next=p->next;??? ????p->next=q;???????????? ????q=p;?????????????????? ????p=head->next;????????? ????}?? ?? ????head?=?q;?? }?? ?? void ?list_destroy(list_node?*head)????{?? ??list_node?*pmove=NULL,*pdel=NULL;?? ??pmove=head;?? ??? ?while (pmove!=head)?? ???{?? ?????pdel=pmove;?? ?????free(pdel);???? ?????pmove=pmove->next;?? ???}?? }?? ?? list_node*?merge_two_list()??? {?? ????list_reverse(list2);?????? ????list_node?*list_merged;???? ????list_node?*p1,*p2,*p0;???????? ????list_merged?=?(list_node*)malloc(sizeof (list_node));?? ????list_merged->data?=?0;?? ????list_merged->next?=?NULL;?? ????p0?=?list_merged;?? ?? ????p1=list1;?? ????p2=list2;?? ????while (p1!=NULL?&&?p2!=NULL)?? ????{?? ????????if (p1->data?<?p2->data)?? ????????{?? ????????????p0->next=(list_node*)malloc(sizeof (list_node));?? ????????????p0->next->data=p1->data;?? ????????????p0->next->next=NULL;?? ????????????p0=p0->next;?? ????????????p1=p1->next;?? ????????}?? ????????else ?? ????????{?? ????????????p0->next=(list_node*)malloc(sizeof (list_node));?? ????????????p0->next->data=p2->data;?? ????????????p0->next->next=NULL;?? ????????????p0=p0->next;?? ????????????p2=p2->next;?? ????????}?? ????}?? ????while (p1!=NULL)?? ????{?? ????????????p0->next=(list_node*)malloc(sizeof (list_node));?? ????????????p0->next->data=p1->data;?? ????????????p0->next->next=NULL;?? ????????????p0=p0->next;?? ????????????p1=p1->next;?? ????}?? ????while (p2!=NULL)?? ????{?? ????????????p0->next=(list_node*)malloc(sizeof (list_node));?? ????????????p0->next->data=p2->data;?? ????????????p0->next->next=NULL;?? ????????????p0=p0->next;?? ????????????p2=p2->next;?? ????}?? ????return ?list_merged;?? }?? ?? ?? ?? list_node*?p0=(list_node*)malloc(sizeof (list_node));?? list_node*?phead=p0;?? list_node*?&p1=list1;??? list_node*?foreach(list_node*?p2)??? {?? ????if (p2==NULL)?return ?phead;?? ?? ????foreach(p2->next);??? ?? ????if (p1->data?>?p2->data)?? ????{?? ????????p0->next?=?(list_node*)malloc(sizeof (list_node));?? ????????p0->next->data?=?p2->data;?? ????????p0->next->next?=?NULL;??? ????????p0=p0->next;?? ????????return ?phead;?? ????}?? ????while (p1!=NULL?&&?p1->data<=p2->data)?? ????{?? ????????p0->next?=?(list_node*)malloc(sizeof (list_node));?? ????????p0->next->data?=?p1->data;?? ????????p0->next->next?=?NULL;??? ????????p0=p0->next;?? ????????p1?=?p1->next;?? ????}?? ????if (p1==NULL)?? ????{?? ????????p0->next?=?(list_node*)malloc(sizeof (list_node));?? ????????p0->next->data?=?p2->data;?? ????????p0->next->next?=?NULL;??? ????????p0=p0->next;?? ????}?? ?? ????return ?phead;?? }?? ?? int ?main()??{?? ????int ?list1_data[]?=?{1,4,6,8,10};?????? ????int ?list2_data[]?=?{14,9,3,2};???????? ?? ????list_create(list1,list1_data,5);????? ????list_create(list2,list2_data,4);????? ????list_print(list1);?? ????list_print(list2);?? ?? ?????? ?????? ?? ????list_node?*list_merged2=foreach(list2);?????? ????list_print(list_merged2->next);?? ?? ????list_destroy(list1);?????????????????? ????list_destroy(list2);?????????????????? ?? ????list_destroy(list_merged2);??????????? ????system("pause" );?? ????return ?0;?? }?? 題 ：如何刪除鏈表的倒數第m的元素？
分析 ： 構建p0,p1兩個迭代器，初始使p0和p1指向頭結點，接著使p1移動到第m+1項，然后指向頭得p0與p1同時前進，當p1指向空節點的時候結束，這時p0所指的位置就是倒數第m個，時間復雜度為O（n）
實現代碼：（ 為了學習的需要，現在C++朋友采用c++實現，不能滿足c朋友要采用c實現的愿望，此實為c++朋友的遺憾，不過一切重在思想。^_^ ）
[cpp] view plaincopy
? ? ? ?? #include?<iostream> ??#include?<list> ??using ?namespace ?std;???? class ?App??{?? ????list<int >?*plist;??? ?? ????void ?delete_node_at_last(int ?m)??? ????{?? ????????list<int >::iterator?p0,?p1;?? ????????p0=p1=p2=plist->begin();?? ?? ?????????? ????????for (int ?i=1;?i<=m;?i++)?? ????????????p1++;?? ?? ?????????? ????????while (p1!=plist->end())?? ????????????p0++,p1++;?? ?? ?????????? ????????plist->erase(p0);?? ????}?? ?? ????void ?create_list()?? ????{?? ????????int ?list_data[]={1,2,3,4,5,6,7,8,9};??? ????????plist?=?new ?list<int >(list_data,?list_data+sizeof (list_data)/sizeof (int ));?? ????}?? ?? ????void ?print_list()?? ????{?? ????????list<int >::iterator?it;?? ????????for (it=plist->begin();?it!=plist->end();?it++)?? ????????????cout<<*it<<'?' ;?? ????}?? ?? public :???????? ????void ?run()?? ????{?? ????????create_list();?? ????????delete_node_at_last(3);??? ????????print_list();?? ????}?? ?? ????~App()?? ????{?? ????????delete ?plist;?? ????}?? };?? ?? int ?main()??{?? App?myapp;?? myapp.run();?? system("pause" );?? return ?0;??}? 題 ：1、如何判斷一個字符串是對稱的？如a，aa，aba。? ?
? ? ? ? ?2、如何利用2函數找出一個字符串中的所有對稱子串？
分析 ：
? ? ? ?? 且 看第一個問題判斷字符串對稱，有以下兩種方法。
? ? ? ? 方法一 、使迭代器p1指向頭，p2指向尾。使p1，p2相對而行(—> | <— ) ，每次比較p1，p2是否相等，直到它們相遇。
? ? ? ??方法二 、使p1、p2指向中間的一個元素（如果字符串長度為偶數的話就指向中間兩個相鄰的元素），使它們相向而行(<— |—> ) ，每次比較p1，p2是否相等。直到它們一個指向頭一個指向尾便結束。??
（可以看出方法1明顯好于方法2）
? ? ? ?? 現 在看第二個問題打印所有的對稱子串，判斷對稱子串的問題我們似乎已經解決，且有以上兩種方法，針對現在的情況是否兩種都合適呢，確切的說不是 。如果是方法一，請想象一下，如果要p1，p2一個指向子串的頭一個指向子串的尾，那么至少要兩層循環一層控制p1一層控制p2，還有一層循環用于判斷子串是否對稱，也就是說總共需要三層嵌套循環，時間復雜度指數為3 。而如果選擇方法二呢？ 兩層循環就能實現，不過要適應現在這種情況需要做些修改，就是針對偶數長度的子串進行一次遍歷，再針對奇數長度的子串進行一次遍歷，也就是兩層嵌套循環中內層有兩次循環，時間復雜度指數為2 。詳情且看代碼。
實現加測試代碼：
[cpp] view plaincopy
? ? ? ?? ?? #include?<iostream> ??#include?<cstdlib> ??#include?<ctime> ??using ?namespace ?std;??class ?App??{?? ????typedef ?string::iterator?Iter_t;?? ?? ????string?m_str;?? ????void ?print(Iter_t?p1,?Iter_t?p2)??? ????{?? ????????while (p1<=p2)?? ????????{?? ????????????cout<<*p1;?? ????????????p1++;?? ????????}?? ????????cout<<"?|?" ;?? ????}?? ????bool ?isHuiwen1(Iter_t?start,Iter_t?end)??? ????{?? ????????Iter_t?p1?=?start;?? ????????Iter_t?p2?=?end;?? ????????int ?times?=?(distance(p1,p2)+1)?/?2;??? ????????while (times)?? ????????{?? ????????????if (*p1?!=?*p2)?? ????????????????return ?false ;?? ?? ????????????p1++;??? ????????????p2--;??? ????????????times--;?? ????????}?? ????????return ?true ;?? ????}?? ?? ????bool ?isHuiwen2(Iter_t?mid_it)??? ????{?? ????????Iter_t?p1,p2;?? ????????p1?=?p2?=?mid_it;?? ?? ????????while (p1>=m_str.begin()?&&?p2?<?m_str.end())??? ????????{????????????? ????????????if (*p1?!=?*p2)??? ????????????????break ;???? ????????????print(p1,p2);?? ????????????p1--,p2++;?? ????????}?? ?? ????????p1?=?p2?=?mid_it;?? ????????p2++;??? ????????while (p1>=m_str.begin()?&&?p2?<?m_str.end())???? ????????{?? ????????????if (*p1?!=?*p2)?? ????????????????return ?false ;?? ????????????print(p1,p2);?? ????????????p1--,p2++;?? ????????}?? ????????return ?true ;?? ????}?? ?? ????void ?show_all_substr_of_huiwen1()??? ????{?? ????????Iter_t?p2=m_str.end()-1;?? ????????Iter_t?p1?=?m_str.begin();?? ????????for (;p1!=m_str.end();p1++)?? ????????????for (p2=p1;p2!=m_str.end();p2++)?? ????????????{?? ????????????????if (isHuiwen1(p1,p2))?? ????????????????????print(p1,p2);?? ????????????}?? ????}?? ?? ????void ?show_all_substr_of_huiwen2()???? ????{?? ????????Iter_t?it?=?m_str.begin();?? ?? ????????for (;it!=m_str.end();it++)?? ????????{?? ????????????isHuiwen2(it);?? ????????}?? ????}?? ?? public :???? ????void ?run()?? ????{?? ????????m_str="abaaba" ;?? ????????cout<<"測試數據為:abaaba" <<endl;?? ????????show_all_substr_of_huiwen1();?? ????????cout<<endl;?? ????????show_all_substr_of_huiwen2();?? ????????cout<<endl;?? ?? ????????m_str="asdfie" ;?? ????????cout<<"測試數據為:asdfie" <<endl;?? ????????show_all_substr_of_huiwen1();?? ????????cout<<endl;?? ????????show_all_substr_of_huiwen2();?? ????????cout<<endl;?? ?? ????????m_str="aabaa" ;?? ????????cout<<"測試數據為:aabaa" <<endl;?? ????????show_all_substr_of_huiwen1();?? ????????cout<<endl;?? ????????show_all_substr_of_huiwen2();?? ????????cout<<endl;?? ?? ?????????? 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;?? ?? ?? ????????time_t ?start,record1;?? ?? ?? ????????cout<<"show?all?substr?of?huiwen?1:" ;?? ????????system("pause" );?? ????????start?=?time(NULL);?????? ????????show_all_substr_of_huiwen1();?? ????????record1?=?time(NULL);?? ?? ????????cout<<endl;?? ????????cout<<"time:" <<record1-start;?? ????????cout<<endl;?? ?? ?? ????????cout<<"show?all?substr?of?huiwen?2:" ;?? ????????system("pause" );?? ????????start?=?time(NULL);?? ????????show_all_substr_of_huiwen2();?? ????????record1?=?time(NULL);?? ?? ????????cout<<endl;?? ????????cout<<"time:" <<record1-start;?? ????????cout<<endl;?? ?? ????????cout<<"(可以看到打印速度明顯快于上一次)" ;?? ????}?? };?? ?? int ?main()??{?? ????App?myapp;?? ????myapp.run();?? ?? ????system("pause" );?? ????return ?0;?? }?? 測試結果：
各測試數據下的一二行為打印出來的所有對稱字串。
按下任意鍵后：
以上是使用方法1打印出來的結果。
按下任意鍵后：
以上便是用方法2打印出來的結果。
題 假設你有一個用1001個整數組成的數組，這些整數是任意排列的，但是你知道所有的整數都在1到1000(包括1000)之間。此外，除一個數字出現兩次外，其他所有數字只出現一次。假設你 只能對這個數組做一次處理 ，用一種算法找出重復的那個數字。 如果你在運算中使用了 輔助的存儲方式，那么你能找到不用這種方式的算法嗎?
分析 ：
方法一 、若 使用輔助的存儲方式，該選擇何種存儲方式呢？ 可使用hash的存儲方式，以1到1000作為hash表的索引，遍歷原數組，統計各數字出現的個數并存儲到以該數字為索引值的hash表中，若某個hash[x]的值為2則退出循環，x就是重復出現兩次的數字。時間復雜度最壞是O(n)。 優點：高效率，缺點：消耗的內存空間過大。 代碼如下：
[cpp] view plaincopy
int ?fun1(const ?int ?a[])??{?? ??int ?hash[1002]={0};?? ??int ?x=0;?? ??for (int ?i?=?0;?i<1001;?i++)?? ????{?? ??????if ((++hash[a[i]])?==?2)?? ????????{?? ??????????x?=?a[i];?? ??????????break ;?? ????????}?? ????}?? ??return ?x;?? }?? 方法二 、若不使用輔助的存儲方式呢？ 已知 1001個整數組成的數組只有一個數字出現了兩次，且整數都在1到1000之間， 所以可推得數組里面包含了1到1000之間的所有數字為[1,2,3……1000]和一個出現兩次的x為1到1000中的任一個數字。這樣就可以計算原數組里的所有數字之和S1和等差數列[1,2,3……1000] 的和 S2，再計算S1與S2之差，該差就是原數組中出現兩次的數字x。時間復雜度是固定的O(n)。優缺點：內存空間消耗幾乎沒有，但是效率要輸于使用hash表的存儲方式。 代碼如下：
[cpp] view plaincopy
int ?fun2(const ?int ?a[])??{?? ??int ?s1=0,s2;?? ??s2?=?1001*1000/2;??? ??for (int ?i?=?0;?i<1001;?i++)?? ????{?? ??????s1+=a[i];?? ????}?? ??return ?s1-s2;?? }?
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