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Update ReplaceSpaces.cpp for array bounds #3

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c125e7e
Update ReplaceSpaces.cpp
wylloong Jul 23, 2017
59b9811
Compiler error: memset was not declared in this scope
wylloong Aug 29, 2017
50525d5
Update StreamMedian.cpp
wylloong Sep 11, 2017
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38 changes: 19 additions & 19 deletions 05_ReplaceSpaces/ReplaceSpaces.cpp
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Original file line number Diff line number Diff line change
Expand Up @@ -8,24 +8,24 @@ Distributed under the BSD license.
*******************************************************************/

//==================================================================
// ����ָOffer�����������Թپ������ͱ���⡷����
// ���ߣ��κ���
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================

// ������5���滻�ո�
// ��Ŀ����ʵ��һ�����������ַ����е�ÿ���ո��滻��"%20"���������롰We are happy.����
// �������We%20are%20happy.����
// 面试题5：替换空格
// 题目：请实现一个函数，把字符串中的每个空格替换成"%20"。例如输入“We are happy.”，
// 则输出“We%20are%20happy.”。

#include <cstdio>
#include <cstring>

/*length Ϊ�ַ�����str�������������ڻ�����ַ���str��ʵ�ʳ���*/
/*length 为字符数组str的总容量，大于或等于字符串str的实际长度*/
void ReplaceBlank(char str[], int length)
{
if(str == nullptr && length <= 0)
return;

/*originalLength Ϊ�ַ���str��ʵ�ʳ���*/
/*originalLength 为字符串str的实际长度*/
int originalLength = 0;
int numberOfBlank = 0;
int i = 0;
Expand All @@ -39,9 +39,9 @@ void ReplaceBlank(char str[], int length)
++ i;
}

/*newLength Ϊ�ѿո��滻��'%20'֮��ij���*/
/*newLength 为把空格替换成'%20'之后的长度*/
int newLength = originalLength + numberOfBlank * 2;
if(newLength > length)
if((newLength+1) > length)
return;

int indexOfOriginal = originalLength;
Expand All @@ -63,7 +63,7 @@ void ReplaceBlank(char str[], int length)
}
}

// ====================���Դ���====================
// ====================测试代码====================
void Test(char* testName, char str[], int length, char expected[])
{
if(testName != nullptr)
Expand All @@ -81,7 +81,7 @@ void Test(char* testName, char str[], int length, char expected[])
printf("failed.\n");
}

// �ո��ھ����м�
// 空格在句子中间
void Test1()
{
const int length = 100;
Expand All @@ -90,7 +90,7 @@ void Test1()
Test("Test1", str, length, "hello%20world");
}

// �ո��ھ��ӿ�ͷ
// 空格在句子开头
void Test2()
{
const int length = 100;
Expand All @@ -99,7 +99,7 @@ void Test2()
Test("Test2", str, length, "%20helloworld");
}

// �ո��ھ���ĩβ
// 空格在句子末尾
void Test3()
{
const int length = 100;
Expand All @@ -108,7 +108,7 @@ void Test3()
Test("Test3", str, length, "helloworld%20");
}

// �����������ո�
// 连续有两个空格
void Test4()
{
const int length = 100;
Expand All @@ -117,13 +117,13 @@ void Test4()
Test("Test4", str, length, "hello%20%20world");
}

// ����nullptr
// 传入nullptr
void Test5()
{
Test("Test5", nullptr, 0, nullptr);
}

// ��������Ϊ�յ��ַ���
// 传入内容为空的字符串
void Test6()
{
const int length = 100;
Expand All @@ -132,7 +132,7 @@ void Test6()
Test("Test6", str, length, "");
}

//��������Ϊһ���ո���ַ���
//传入内容为一个空格的字符串
void Test7()
{
const int length = 100;
Expand All @@ -141,7 +141,7 @@ void Test7()
Test("Test7", str, length, "%20");
}

// ������ַ���û�пո�
// 传入的字符串没有空格
void Test8()
{
const int length = 100;
Expand All @@ -150,7 +150,7 @@ void Test8()
Test("Test8", str, length, "helloworld");
}

// ������ַ���ȫ�ǿո�
// 传入的字符串全是空格
void Test9()
{
const int length = 100;
Expand Down
27 changes: 14 additions & 13 deletions 17_Print1ToMaxOfNDigits/Print1ToMaxOfNDigits.cpp
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Original file line number Diff line number Diff line change
Expand Up @@ -8,22 +8,23 @@ Distributed under the BSD license.
*******************************************************************/

//==================================================================
// ����ָOffer�����������Թپ������ͱ���⡷����
// ���ߣ��κ���
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================

// ������17����ӡ1������nλ��
// ��Ŀ����������n����˳���ӡ����1����nλʮ����������������3����
// ��ӡ��1��2��3һֱ������3λ����999��
// 面试题17：打印1到最大的n位数
// 题目：输入数字n，按顺序打印出从1最大的n位十进制数。比如输入3，则
// 打印出1、2、3一直到最大的3位数即999。

#include <cstdio>
#include <memory>
#include <cstring>

void PrintNumber(char* number);
bool Increment(char* number);
void Print1ToMaxOfNDigitsRecursively(char* number, int length, int index);

// ====================����һ====================
// ====================方法一====================
void Print1ToMaxOfNDigits_1(int n)
{
if (n <= 0)
Expand All @@ -41,8 +42,8 @@ void Print1ToMaxOfNDigits_1(int n)
delete[]number;
}

// �ַ���number��ʾһ�����֣��� number������1
// ������ӷ�������򷵻�true������Ϊfalse
// 字符串number表示一个数字，在 number上增加1
// 如果做加法溢出，则返回true；否则为false
bool Increment(char* number)
{
bool isOverflow = false;
Expand Down Expand Up @@ -76,7 +77,7 @@ bool Increment(char* number)
return isOverflow;
}

// ====================������====================
// ====================方法二====================
void Print1ToMaxOfNDigits_2(int n)
{
if (n <= 0)
Expand Down Expand Up @@ -109,9 +110,9 @@ void Print1ToMaxOfNDigitsRecursively(char* number, int length, int index)
}
}

// ====================��������====================
// �ַ���number��ʾһ�����֣����������ɸ�0��ͷ
// ��ӡ��������֣������Կ�ͷ��0
// ====================公共函数====================
// 字符串number表示一个数字，数字有若干个0开头
// 打印出这个数字，并忽略开头的0
void PrintNumber(char* number)
{
bool isBeginning0 = true;
Expand All @@ -131,7 +132,7 @@ void PrintNumber(char* number)
printf("\t");
}

// ====================���Դ���====================
// ====================测试代码====================
void Test(int n)
{
printf("Test for %d begins:\n", n);
Expand Down
16 changes: 8 additions & 8 deletions 41_StreamMedian/StreamMedian.cpp
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Original file line number Diff line number Diff line change
Expand Up @@ -8,14 +8,14 @@ Distributed under the BSD license.
*******************************************************************/

//==================================================================
// ����ָOffer�����������Թپ������ͱ���⡷����
// ���ߣ��κ���
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================

// ������41���������е���λ��
// ��Ŀ����εõ�һ���������е���λ����������������ж�����������ֵ����ô
// ��λ������������ֵ����֮��λ���м����ֵ��������������ж���ż������ֵ��
// ��ô��λ������������ֵ����֮���м���������ƽ��ֵ��
// 面试题41：数据流中的中位数
// 题目：如何得到一个数据流中的中位数？如果从数据流中读出奇数个数值，那么
// 中位数就是所有数值排序之后位于中间的数值。如果从数据流中读出偶数个数值，
// 那么中位数就是所有数值排序之后中间两个数的平均值。

#include <cstdio>
#include <algorithm>
Expand Down Expand Up @@ -73,7 +73,7 @@ template<typename T> class DynamicArray
if((size & 1) == 1)
median = min[0];
else
median = (min[0] + max[0]) / 2;
median = (min[0] + max[0]) / 2.0;

return median;
}
Expand All @@ -83,7 +83,7 @@ template<typename T> class DynamicArray
vector<T> max;
};

// ====================���Դ���====================
// ====================测试代码====================
void Test(char* testName, DynamicArray<double>& numbers, double expected)
{
if(testName != nullptr)
Expand Down