Practice in program

1. 将一个十进制整数转换成任意进制（2-36）的整数：

#include <cstdio>#include <cstdlib>int Tentrans2any(char* out, int len, int number, int base) {if( !out || len < 2 || base < 2 || base > 36 ) return 0;bool neg = number < 0 ? true : false;int index = 0;if ( neg ) {out[index++] = '-';number = -number;}do {int digit = number % base;if ( digit > 9 ) {out[index++] = digit - 10 + 'A';} else {out[index++] = digit + '0';}number /= base;} while ( index < len && number != 0 );// reversint li = neg ? 1:0;int ri = index-1;int temp = 0;while ( li < ri ) {temp = out[li];out[li] = out[ri];out[ri] = temp;li++;ri--;}return index;}int main(int argc, char** argv){if ( argc < 3 ) {printf("Usage:\n%s number base\n", argv[0]);return 1;}int number = strtol(argv[1], NULL, 10);int base = strtol(argv[2], NULL, 10);char out[128] = {0};int ret = Tentrans2any( out, sizeof(out), number, base);printf("Ret %d, Number %d in %d = %s\n", ret, number, base, out);return 0;}

2. 位移代替乘除法

a*10 = a*8 + a*2 = a<<3 + a<1;

a / 8 = a >> 3; // 除法暂时没找到非2^n次的位移方法，比如a/10。

另外，编译器在处理a * 2^n 或 a / 2^n 时会做优化的位移处理。这个Stackoverflow给出了乘法的汇编解析。

下面这个是 12/2 的汇编过程（VS2005：cl.exe 14.00.50727.42 for 80x86），也是类似的处理。sar eax,1 即是右移1位的意思。

  int i = 12;00401014  mov         dword ptr [i],0Ch   i = i / 2;0040101B  mov         eax,dword ptr [i] 0040101E  cdq              0040101F  sub         eax,edx 00401021  sar         eax,1 00401023  mov         dword ptr [i],eax 

自己另外写了个位移的乘法小程序，并没有什么卵用。

#include <cstdio>#include <cstdlib>#include <string.h>int main(int argc, char** argv) {if ( argc < 2 ) {printf("Usage:\n%s number * number\n", argv[0]);return 1;}char line[128] = {0};for( int i=1; i<argc; ++i ) {strncat( line, argv[i], sizeof(line)-strlen(line)-1 );}char *pEnd = NULL;long int n1 = strtol( line, &pEnd, 10);bool negitive = false;if ( n1 < 0 ) {n1 = -n1;negitive = true;}while( pEnd && *pEnd != 0 && *pEnd != '*' )pEnd++;if ( *pEnd != '*' ) {printf("Wrong args\n");return 2;}long int n2 = strtol( pEnd+1, NULL, 10 );if ( n2 < 0 ) {n2 = -n2;negitive = negitive ? false:true;}long int result = 0;if ( n1 != 0 && n2 != 0 ) {int bits = sizeof(result)*8;if ( (n2&0x01) ) result += n1;for ( int i=1; i<bits; ++i ) {if ( (n2&(0x01<<i)) != 0 ) {result += (n1 << i);}}if ( negitive ) result = - result;}printf("%s = %d\n", line, result);return 0;}

3. 合并两个有序的链表，下面的算法思路是在一次畅游的面试过程中得知的：既然都是有序的链表，那么就可以两个链表同时遍历，通过转移next指针来完成两个链表的完全合并。

#include <cstdio>#include <cstdlib>#include <ctime>struct Node{int data;Node* next;};Node* List1 = NULL;Node* List2 = NULL;Node* GenerateList(int count) {if ( count < 1 ) return NULL;static int tkey = 0x5241;Node head;Node* pHead = &head;srand( time(NULL)+tkey++ );int begin = rand() % count;int actcount = 0;for( int i=begin; actcount<count; ++i ) {if ( rand()%3 == 0 ) {Node *pn = new Node;pn->data = i;pn->next = NULL;pHead->next = pn;pHead = pn;actcount++;}}return head.next;}void PrintList( Node* list ) {int count = 0;while ( list ) {printf( "%d ", list->data );list = list->next;count ++;}printf(":(%d)\n", count);}Node* MergeSeqlists(Node* list1, Node* list2){if ( list1 == NULL ) return list2;if ( list2 == NULL ) return list1;// set headNode* head = list1->data <= list2->data ? list1:list2;while (list1 && list2 ){if ( list1->data <= list2->data ) {while ( list1->next && list1->next->data <= list2->data ) {list1 = list1->next;continue;}Node* temp = list1->next;list1->next = list2;list1 = temp;} else {while ( list2->next && list2->next->data <= list1->data ) {list2 = list2->next;continue;}Node* temp = list2->next;list2->next = list1;list2 = temp;}}return head;}int main(int argc, char** argv) {if ( argc < 3 ) {printf("Usage:\n%s list1counts list2counts\n", argv[1]);return 1;}int count = strtol( argv[1], NULL, 10 );List1 = GenerateList( count );count = strtol( argv[2], NULL, 10 );List2 = GenerateList( count );PrintList( List1 );PrintList( List2 );Node* list = MergeSeqlists( List1, List2 );PrintList( list );return 0;}

另外在剑指offer上面看到了个递归的实现方式

Node* MergeSeqlists2(Node* list1, Node* list2) {if ( list1 == NULL ) return list2;if ( list2 == NULL ) return list1;Node* pHead = NULL;if ( list1->data < list2->data ) {pHead = list1;pHead->next = MergeSeqlists2( list1->next, list2 );} else {pHead = list2;pHead->next = MergeSeqlists2( list1, list2->next );}return pHead;}