career_cup
2013-06-05 09:46
302 查看
[1]反转字符串
#include "stdafx.h"
void reverse(char *str) {
char *end =str;
char tmp;
if(str){
while(*end){
end++;
}
end--;
while(str<end){
tmp= *str;
*str++=*end;
*end--=tmp;
}
}
}
int main(int argc, char* argv[])
{
char testStr[10] = "abcdefg";
reverse(testStr);
printf("%s\n",testStr);
return 0;
}
[2]将字符数组相同的字符去掉
public class CharDemo1{
public static void main(String args[]){
String str1 = "abcdabfh";
char testStr[] = str1.toCharArray();
removeDuplicates(testStr);
for(int i=0;i<testStr.length;i++){
System.out.print(testStr[i]);
}
}
public static void removeDuplicates(char[] str) {
if (str == null) return;
int len = str.length;
if (len < 2) return;
int tail = 1;
for (int i = 1; i < len; i++) {
int j;
for (j = 0; j < tail; j++) {
if (str[i] == str[j]) break;
}
if (j == tail) {
str[tail] = str[i];
tail++;
}
}
for(int i= tail;i<len;i++){
str[i] = 0;
}
}
}
输出：abcdfh
[3]获取当前时间
#include "stdafx.h"
#include "stdio.h"
#include "stdlib.h"
#include <string.h>
#include <time.h>
#include <windows.h>
int main(int argc, char* argv[])
{
FILE *fp;
int line = 0;
char buf[Num];
if((fp=fopen("d:/time.txt","a"))==NULL){
fprintf(stderr,"don't fopen %s=\n",strerror(errno));
return -1;
}
while(fgets(buf,Num,fp)!=NULL){
if(strlen(buf)<Num-1 ||buf[Num-2]!='\n')
line++;
}
while(1)
{
time_t t;
time(&t);
struct tm *t1;
t1 = localtime(&t);
sprintf(buf,"%d,%d,%d,%d,%d,%d,%d\n",++line,t1->tm_year+1900,
t1->tm_mon,t1->tm_mday,t1->tm_hour,t1->tm_min,t1->tm_sec);
printf("the current time is =%s\n",buf);
fputs(buf,fp);
fflush(fp);
Sleep(1);
}
fclose(fp);
return 0;
}
[4]
int a=10;
int b;
int main(int argc, char* argv[])
{
int i;
printf("i=%d\n",i);
printf("&i=%d\n",&i);
for(i=0;i<a;i++)
b+=3*i;
for(i=0;i<10;i++,a++)
{
;
}
printf("a=%d\n",a);
printf("b=%d\n",b);
return 0;
}
[5]
int foo()
{
int a =0;
a=strlen("aaabbbccc")-10;
return a;
}
int main(int argc, char* argv[])
{
unsigned char a =foo();
printf("a =%d\n",a);
return 0;
}
a =255
[6]
int main(int argc, char* argv[])
{
unsigned char a = -1;
unsigned char b = 257;
unsigned char c = (a*b+4)/2;
printf("c =%d\n",c);
return 0;
}
c=129
[7]
#define FREE(p) free(p);\
p = NULL;
void freeFunc(char *p)
{
free(p);
p=NULL;
}
int main(int argc, char* argv[])
{
char * buf = (char *)malloc(100);
freeFunc(buf);
//FREE(buf);
return 0;
}
freeFunc(char *p)这种方法只能释放内存，但buf并不会=NULL;并不能做到无残留
而FREE(p)则可以。
[8]
#define FREE(p) free(p);\
p = NULL;
int main(int argc, char* argv[])
{
int tag=2;
int i;
char * buf = (char *)malloc(100);
if(tag==3)
FREE(buf);
for(i=0;i<10;i++)
{
buf[i]=2*i+'a';
}
buf[i]=0;
printf("buf is %s\n",buf);
return 0;
}
这段程序会崩溃
[9]
#include "stdafx.h"
#include<iostream.h>
#include<string.h>
#include <malloc.h>
#include <stdio.h>
#include <stdlib.h>
#include <memory.h>
#define F_MM(f, m, n) \
{ \
(f), \
(m), \
~((n)-(m)) * !!(n), \
~(n),\
}
struct clk_freq_tbl {
unsigned long freq_hz;
int m_val;
int n_val;
int d_val;
};
static struct clk_freq_tbl ftbl_csi0_1_clk[] = {
F_MM(100000000, 1, 5),
//F_END,
};
int main(int argc, char* argv[])
{
printf("freq is =%d,m_val is =%d,n_val is =%d,d_val is =%d\n",ftbl_csi0_1_clk[0].freq_hz,ftbl_csi0_1_clk[0].m_val,\
ftbl_csi0_1_clk[0].n_val,ftbl_csi0_1_clk[0].d_val);
return 0;
}
[10]
转载请注明出处：/article/1378883.html
题目：
You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1’s digit is at the head of the list. Write a function that
adds the two numbers and returns the sum as a linked list.
EXAMPLE
Input: (3 -> 1 -> 5), (5 -> 9 -> 2)
Output: 8 -> 0 -> 8
翻译：
用单链表表示一个正整数，每个结点代表其中的一位数字，逆序排列，个位位于表头，最高位位于表尾，将两个数相加并返回结果，结果也由链表表示。
例如：
输入：(3 -> 1 -> 5)+(5 -> 9 -> 2)
输出：8 -> 0 -> 8
思路：
这道题目不难，也没有太大的技巧性，就按照最常规的思路来，只是要注意将所有的情况全部考虑到。
1、两个链表中有一个为NULL，这时直接返回另一个链表就行了；
2、如果两个链表都为空，这本身也包含在第1种情况中包；
3、如果两个链表长度不等，我们需要将二者相加后的结果保存在较长的链表上；
4、如果某位相加大于等于10，需要进位；
5、如果相加后的链表长度大于另两个链表中最长的那个链表的长度，则需要开辟新的节点，将其挂在长度较长的那个链表的表尾。
实际上该题目也为我们提供了一种实现大数相加的思路。
实现代码：
#include <stdio.h>
#include <stdlib.h>
#define bool int
#define true 1
#define false 0
typedef struct Node
{
int data;
struct Node *pNext;
}NODE,*PNODE;
PNODE create_list();
void traverse_list(PNODE);
bool lct_empty(PNODE);
int length_list(PNODE);
void clear_list(PNODE);
PNODE addList(PNODE,PNODE);
void AddShortToLong(PNODE,PNODE);
int main(int argc,char *argv[])
{
int c;
printf("Create the first list:\n");
PNODE pHead1 = create_list();
printf("List 1:\n");
traverse_list(pHead1);
//fflush(stdin); //刷新输入缓冲区
while((c=getchar())!='\n'&&c!=EOF);
printf("Create the second list:\n");
PNODE pHead2 = create_list();
printf("List 2:\n");
traverse_list(pHead2);
PNODE pHead = addList(pHead1,pHead2);
printf("After added,the new List:\n");
traverse_list(pHead);
return 0;
}
/*
创建一个链表,并返回头指针
*/
PNODE create_list()
{
int val;
PNODE pHead =(PNODE)malloc(sizeof(NODE));
PNODE pCurrent = pHead;
pCurrent->pNext = NULL;
if(NULL == pHead)
{
printf("pHead malloc failed!");
exit(-1);
}
printf("Input first data(q to quit):");
while(scanf("%d",&val)==1)
{
PNODE pNew = (PNODE)malloc(sizeof(NODE));
if(NULL == pNew)
{
printf("pNew malloc failed!");
exit(-1);
}
pNew->data = val;
pCurrent->pNext = pNew;
pNew->pNext = NULL;
pCurrent = pNew;
printf("Input next data(q to quit):");
}
return pHead;
}
/*
遍历链表
*/
void traverse_list(PNODE pHead)
{
PNODE pCurrent = pHead->pNext;
printf("now dataes in the list are:\n");
while(pCurrent != NULL)
{
printf("%d ",pCurrent->data);
pCurrent = pCurrent->pNext;
}
printf("\n");
return;
}
/*
判断链表是否为空
*/
bool lct_empty(PNODE pNode)
{
if(NULL == pNode->pNext)
return true;
else
return false;
}
/*
求链表长度，即节点总数（不计入头节点）
*/
int length_list(PNODE pNode)
{
int count = 0;
PNODE pCurrent = pNode->pNext;
while(pCurrent != NULL)
{
count++;
pCurrent = pCurrent->pNext;
}
return count;
}
/*
清空链表，即使链表只剩下头节点（头节点中没有数据）
*/
void clear_list(PNODE pHead)
{
PNODE p = pHead->pNext;
PNODE r = NULL;
while(p != NULL)
{
r = p->pNext;
free(p);
p = r;
}
pHead->pNext = NULL;
return ;
}
/*
两个链表相加
*/
PNODE addList(PNODE pHead1,PNODE pHead2)
{
if(!pHead1)
return pHead2;
if(!pHead2)
return pHead1;
int len1 = length_list(pHead1);
int len2 = length_list(pHead2);
if(len1 >= len2)
{
AddShortToLong(pHead1,pHead2);
return pHead1;
}
else
{
AddShortToLong(pHead2,pHead1);
return pHead2;
}
}
/*
第一个链表的长度大于或等于第二个链表的长度，
将第二个链表加到第一个链表上
*/
void AddShortToLong(PNODE pHeadLong,PNODE pHeadShort)
{
PNODE p1 = pHeadLong->pNext;
PNODE p2 = pHeadShort->pNext;
while(p1 && p2)
{
p1->data = p1->data + p2->data;
if(p1->data >= 10)
{
p1->data -= 10;
if(p1->pNext)
{
p1->pNext->data++;
if(p1->pNext->data >= 10) //两链表长度不同，最后一位又有进位
{
p1->pNext->data -= 10;
PNODE pNew = (PNODE)malloc(sizeof(NODE));
if(!pNew)
{
printf("malloc failed\n");
exit(-1);
}
pNew->pNext = NULL;
pNew->data = 1;
p1->pNext->pNext = pNew;
}
}
else //两链表长度相同，且组后一位有进位
{
PNODE pNew = (PNODE)malloc(sizeof(NODE));
if(!pNew)
{
printf("malloc failed\n");
exit(-1);
}
pNew->pNext = NULL;
pNew->data = 1;
p1->pNext = pNew;
}
}
p1 = p1->pNext;
p2 = p2->pNext;
}
}
#include "stdafx.h"
void reverse(char *str) {
char *end =str;
char tmp;
if(str){
while(*end){
end++;
}
end--;
while(str<end){
tmp= *str;
*str++=*end;
*end--=tmp;
}
}
}
int main(int argc, char* argv[])
{
char testStr[10] = "abcdefg";
reverse(testStr);
printf("%s\n",testStr);
return 0;
}
[2]将字符数组相同的字符去掉
public class CharDemo1{
public static void main(String args[]){
String str1 = "abcdabfh";
char testStr[] = str1.toCharArray();
removeDuplicates(testStr);
for(int i=0;i<testStr.length;i++){
System.out.print(testStr[i]);
}
}
public static void removeDuplicates(char[] str) {
if (str == null) return;
int len = str.length;
if (len < 2) return;
int tail = 1;
for (int i = 1; i < len; i++) {
int j;
for (j = 0; j < tail; j++) {
if (str[i] == str[j]) break;
}
if (j == tail) {
str[tail] = str[i];
tail++;
}
}
for(int i= tail;i<len;i++){
str[i] = 0;
}
}
}
输出：abcdfh
[3]获取当前时间
#include "stdafx.h"
#include "stdio.h"
#include "stdlib.h"
#include <string.h>
#include <time.h>
#include <windows.h>
int main(int argc, char* argv[])
{
FILE *fp;
int line = 0;
char buf[Num];
if((fp=fopen("d:/time.txt","a"))==NULL){
fprintf(stderr,"don't fopen %s=\n",strerror(errno));
return -1;
}
while(fgets(buf,Num,fp)!=NULL){
if(strlen(buf)<Num-1 ||buf[Num-2]!='\n')
line++;
}
while(1)
{
time_t t;
time(&t);
struct tm *t1;
t1 = localtime(&t);
sprintf(buf,"%d,%d,%d,%d,%d,%d,%d\n",++line,t1->tm_year+1900,
t1->tm_mon,t1->tm_mday,t1->tm_hour,t1->tm_min,t1->tm_sec);
printf("the current time is =%s\n",buf);
fputs(buf,fp);
fflush(fp);
Sleep(1);
}
fclose(fp);
return 0;
}
[4]
int a=10;
int b;
int main(int argc, char* argv[])
{
int i;
printf("i=%d\n",i);
printf("&i=%d\n",&i);
for(i=0;i<a;i++)
b+=3*i;
for(i=0;i<10;i++,a++)
{
;
}
printf("a=%d\n",a);
printf("b=%d\n",b);
return 0;
}
[5]
int foo()
{
int a =0;
a=strlen("aaabbbccc")-10;
return a;
}
int main(int argc, char* argv[])
{
unsigned char a =foo();
printf("a =%d\n",a);
return 0;
}
a =255
[6]
int main(int argc, char* argv[])
{
unsigned char a = -1;
unsigned char b = 257;
unsigned char c = (a*b+4)/2;
printf("c =%d\n",c);
return 0;
}
c=129
[7]
#define FREE(p) free(p);\
p = NULL;
void freeFunc(char *p)
{
free(p);
p=NULL;
}
int main(int argc, char* argv[])
{
char * buf = (char *)malloc(100);
freeFunc(buf);
//FREE(buf);
return 0;
}
freeFunc(char *p)这种方法只能释放内存，但buf并不会=NULL;并不能做到无残留
而FREE(p)则可以。
[8]
#define FREE(p) free(p);\
p = NULL;
int main(int argc, char* argv[])
{
int tag=2;
int i;
char * buf = (char *)malloc(100);
if(tag==3)
FREE(buf);
for(i=0;i<10;i++)
{
buf[i]=2*i+'a';
}
buf[i]=0;
printf("buf is %s\n",buf);
return 0;
}
这段程序会崩溃
[9]
#include "stdafx.h"
#include<iostream.h>
#include<string.h>
#include <malloc.h>
#include <stdio.h>
#include <stdlib.h>
#include <memory.h>
#define F_MM(f, m, n) \
{ \
(f), \
(m), \
~((n)-(m)) * !!(n), \
~(n),\
}
struct clk_freq_tbl {
unsigned long freq_hz;
int m_val;
int n_val;
int d_val;
};
static struct clk_freq_tbl ftbl_csi0_1_clk[] = {
F_MM(100000000, 1, 5),
//F_END,
};
int main(int argc, char* argv[])
{
printf("freq is =%d,m_val is =%d,n_val is =%d,d_val is =%d\n",ftbl_csi0_1_clk[0].freq_hz,ftbl_csi0_1_clk[0].m_val,\
ftbl_csi0_1_clk[0].n_val,ftbl_csi0_1_clk[0].d_val);
return 0;
}
[10]
转载请注明出处：/article/1378883.html
题目：
You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1’s digit is at the head of the list. Write a function that
adds the two numbers and returns the sum as a linked list.
EXAMPLE
Input: (3 -> 1 -> 5), (5 -> 9 -> 2)
Output: 8 -> 0 -> 8
翻译：
用单链表表示一个正整数，每个结点代表其中的一位数字，逆序排列，个位位于表头，最高位位于表尾，将两个数相加并返回结果，结果也由链表表示。
例如：
输入：(3 -> 1 -> 5)+(5 -> 9 -> 2)
输出：8 -> 0 -> 8
思路：
这道题目不难，也没有太大的技巧性，就按照最常规的思路来，只是要注意将所有的情况全部考虑到。
1、两个链表中有一个为NULL，这时直接返回另一个链表就行了；
2、如果两个链表都为空，这本身也包含在第1种情况中包；
3、如果两个链表长度不等，我们需要将二者相加后的结果保存在较长的链表上；
4、如果某位相加大于等于10，需要进位；
5、如果相加后的链表长度大于另两个链表中最长的那个链表的长度，则需要开辟新的节点，将其挂在长度较长的那个链表的表尾。
实际上该题目也为我们提供了一种实现大数相加的思路。
实现代码：
#include <stdio.h>
#include <stdlib.h>
#define bool int
#define true 1
#define false 0
typedef struct Node
{
int data;
struct Node *pNext;
}NODE,*PNODE;
PNODE create_list();
void traverse_list(PNODE);
bool lct_empty(PNODE);
int length_list(PNODE);
void clear_list(PNODE);
PNODE addList(PNODE,PNODE);
void AddShortToLong(PNODE,PNODE);
int main(int argc,char *argv[])
{
int c;
printf("Create the first list:\n");
PNODE pHead1 = create_list();
printf("List 1:\n");
traverse_list(pHead1);
//fflush(stdin); //刷新输入缓冲区
while((c=getchar())!='\n'&&c!=EOF);
printf("Create the second list:\n");
PNODE pHead2 = create_list();
printf("List 2:\n");
traverse_list(pHead2);
PNODE pHead = addList(pHead1,pHead2);
printf("After added,the new List:\n");
traverse_list(pHead);
return 0;
}
/*
创建一个链表,并返回头指针
*/
PNODE create_list()
{
int val;
PNODE pHead =(PNODE)malloc(sizeof(NODE));
PNODE pCurrent = pHead;
pCurrent->pNext = NULL;
if(NULL == pHead)
{
printf("pHead malloc failed!");
exit(-1);
}
printf("Input first data(q to quit):");
while(scanf("%d",&val)==1)
{
PNODE pNew = (PNODE)malloc(sizeof(NODE));
if(NULL == pNew)
{
printf("pNew malloc failed!");
exit(-1);
}
pNew->data = val;
pCurrent->pNext = pNew;
pNew->pNext = NULL;
pCurrent = pNew;
printf("Input next data(q to quit):");
}
return pHead;
}
/*
遍历链表
*/
void traverse_list(PNODE pHead)
{
PNODE pCurrent = pHead->pNext;
printf("now dataes in the list are:\n");
while(pCurrent != NULL)
{
printf("%d ",pCurrent->data);
pCurrent = pCurrent->pNext;
}
printf("\n");
return;
}
/*
判断链表是否为空
*/
bool lct_empty(PNODE pNode)
{
if(NULL == pNode->pNext)
return true;
else
return false;
}
/*
求链表长度，即节点总数（不计入头节点）
*/
int length_list(PNODE pNode)
{
int count = 0;
PNODE pCurrent = pNode->pNext;
while(pCurrent != NULL)
{
count++;
pCurrent = pCurrent->pNext;
}
return count;
}
/*
清空链表，即使链表只剩下头节点（头节点中没有数据）
*/
void clear_list(PNODE pHead)
{
PNODE p = pHead->pNext;
PNODE r = NULL;
while(p != NULL)
{
r = p->pNext;
free(p);
p = r;
}
pHead->pNext = NULL;
return ;
}
/*
两个链表相加
*/
PNODE addList(PNODE pHead1,PNODE pHead2)
{
if(!pHead1)
return pHead2;
if(!pHead2)
return pHead1;
int len1 = length_list(pHead1);
int len2 = length_list(pHead2);
if(len1 >= len2)
{
AddShortToLong(pHead1,pHead2);
return pHead1;
}
else
{
AddShortToLong(pHead2,pHead1);
return pHead2;
}
}
/*
第一个链表的长度大于或等于第二个链表的长度，
将第二个链表加到第一个链表上
*/
void AddShortToLong(PNODE pHeadLong,PNODE pHeadShort)
{
PNODE p1 = pHeadLong->pNext;
PNODE p2 = pHeadShort->pNext;
while(p1 && p2)
{
p1->data = p1->data + p2->data;
if(p1->data >= 10)
{
p1->data -= 10;
if(p1->pNext)
{
p1->pNext->data++;
if(p1->pNext->data >= 10) //两链表长度不同，最后一位又有进位
{
p1->pNext->data -= 10;
PNODE pNew = (PNODE)malloc(sizeof(NODE));
if(!pNew)
{
printf("malloc failed\n");
exit(-1);
}
pNew->pNext = NULL;
pNew->data = 1;
p1->pNext->pNext = pNew;
}
}
else //两链表长度相同，且组后一位有进位
{
PNODE pNew = (PNODE)malloc(sizeof(NODE));
if(!pNew)
{
printf("malloc failed\n");
exit(-1);
}
pNew->pNext = NULL;
pNew->data = 1;
p1->pNext = pNew;
}
}
p1 = p1->pNext;
p2 = p2->pNext;
}
}
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