20、【剑指Offer学习】【面试题20：表示数值的字符串】
46、【剑指Offer学习】【面试题46：把数字翻译成字符串】
48、【剑指Offer学习】【面试题48：最长不含重复字符的子字符串】
50、【剑指Offer学习】【面试题50：字符串中第一个只出现一次的字符】
67、【剑指Offer学习】【面试题67：把字符串转换成整数】
20. 表示数值的字符串
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题20：表示数值的字符串
// 题目：请实现一个函数用来判断字符串是否表示数值（包括整数和小数）。例如，
// 字符串“+100”、“5e2”、“-123”、“3.1416”及“-1E-16”都表示数值，但“12e”、
// “1a3.14”、“1.2.3”、“+-5”及“12e+5.4”都不是
#include <stdio.h>
bool scanUnsignedInteger(const char** str);
bool scanInteger(const char** str);
// 数字的格式可以用A[.[B]][e|EC]或者.B[e|EC]表示，其中A和C都是
// 整数（可以有正负号，也可以没有），而B是一个无符号整数
bool isNumeric(const char* str)
{
if(str == nullptr)
return false;
bool numeric = scanInteger(&str);
// 如果出现'.'，接下来是数字的小数部分
if(*str == '.')
{
++str;
// 下面一行代码用||的原因：
// 1. 小数可以没有整数部分，例如.123等于0.123；
// 2. 小数点后面可以没有数字，例如233.等于233.0；
// 3. 当然小数点前面和后面可以有数字，例如233.666
numeric = scanUnsignedInteger(&str) || numeric;
}
// 如果出现'e'或者'E'，接下来跟着的是数字的指数部分
if(*str == 'e' || *str == 'E')
{
++str;
// 下面一行代码用&&的原因：
// 1. 当e或E前面没有数字时，整个字符串不能表示数字，例如.e1、e1；
// 2. 当e或E后面没有整数时，整个字符串不能表示数字，例如12e、12e+5.4
numeric = numeric && scanInteger(&str);
}
return numeric && *str == '\0';
}
bool scanUnsignedInteger(const char** str)
{
const char* before = *str;
while(**str != '\0' && **str >= '0' && **str <= '9') ++(*str); 当str中存在若干0-9的数字时，返回true return *str> before;
}
// 整数的格式可以用[+|-]B表示, 其中B为无符号整数
bool scanInteger(const char** str)
{
if(**str == '+' || **str == '-')
++(*str);
return scanUnsignedInteger(str);
}
// ====================测试代码====================
void Test(const char* testName, const char* str, bool expected)
{
if(testName != nullptr)
printf("%s begins: ", testName);
if(isNumeric(str) == expected)
printf("Passed.\n");
else
printf("FAILED.\n");
}
int main(int argc, char* argv[])
{
Test("Test1", "100", true);
Test("Test2", "123.45e+6", true);
Test("Test3", "+500", true);
Test("Test4", "5e2", true);
Test("Test5", "3.1416", true);
Test("Test6", "600.", true);
Test("Test7", "-.123", true);
Test("Test8", "-1E-16", true);
Test("Test9", "1.79769313486232E+308", true);
printf("\n\n");
Test("Test10", "12e", false);
Test("Test11", "1a3.14", false);
Test("Test12", "1+23", false);
Test("Test13", "1.2.3", false);
Test("Test14", "+-5", false);
Test("Test15", "12e+5.4", false);
Test("Test16", ".", false);
Test("Test17", ".e1", false);
Test("Test18", "e1", false);
Test("Test19", "+.", false);
Test("Test20", "", false);
Test("Test21", nullptr, false);
return 0;
}
</=></stdio.h>
38. 面试题38：字符串的排列
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题38：字符串的排列
// 题目：输入一个字符串，打印出该字符串中字符的所有排列。例如输入字符串abc，
// 则打印出由字符a、b、c所能排列出来的所有字符串abc、acb、bac、bca、cab和cba。
#include <cstdio>
void Permutation(char* pStr, char* pBegin);
void Permutation(char* pStr)
{
if(pStr == nullptr)
return;
Permutation(pStr, pStr);
}
void Permutation(char* pStr, char* pBegin)
{
if(*pBegin == '\0')
{
printf("%s\n", pStr);
}
else
{
for(char* pCh = pBegin; *pCh != '\0'; ++ pCh)
{
char temp = *pCh;
*pCh = *pBegin;
*pBegin = temp;
Permutation(pStr, pBegin + 1);
temp = *pCh;
*pCh = *pBegin;
*pBegin = temp;
}
}
}
// ====================测试代码====================
void Test(char* pStr)
{
if(pStr == nullptr)
printf("Test for nullptr begins:\n");
else
printf("Test for %s begins:\n", pStr);
Permutation(pStr);
printf("\n");
}
int main(int argc, char* argv[])
{
Test(nullptr);
char string1[] = "";
Test(string1);
char string2[] = "a";
Test(string2);
char string3[] = "ab";
Test(string3);
char string4[] = "abc";
Test(string4);
return 0;
}
</cstdio>
46. 面试题46：把数字翻译成字符串
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题46：把数字翻译成字符串
// 题目：给定一个数字，我们按照如下规则把它翻译为字符串：0翻译成"a"，1翻
// 译成"b"，……，11翻译成"l"，……，25翻译成"z"。一个数字可能有多个翻译。例
// 如12258有5种不同的翻译，它们分别是"bccfi"、"bwfi"、"bczi"、"mcfi"和
// "mzi"。请编程实现一个函数用来计算一个数字有多少种不同的翻译方法。
#include <string>
#include <iostream>
using namespace std;
int GetTranslationCount(const string& number);
int GetTranslationCount(int number)
{
if(number < 0)
return 0;
string numberInString = to_string(number);
return GetTranslationCount(numberInString);
}
int GetTranslationCount(const string& number)
{
int length = number.length();
int* counts = new int[length];
int count = 0;
for(int i = length - 1; i >= 0; --i)
{
count = 0;
if(i < length - 1)
count = counts[i + 1];
else
count = 1;
if(i < length - 1)
{
int digit1 = number[i] - '0';
int digit2 = number[i + 1] - '0';
int converted = digit1 * 10 + digit2;
if(converted >= 10 && converted <= 25) { if(i < length - 2) count +="counts[i" 2]; else } counts[i]="count;" delete[] counts; return count; =="==================测试代码====================" void test(const string& testname, int number, expected) if(gettranslationcount(number)="=" cout << testname " passed." endl; failed." test1() number="0;" expected="1;" test("test1", expected); test2() test("test2", test3() test("test3", test4() test("test4", test5() test("test5", test6() test("test6", test7() test("test7", test8() test("test8", test9() test("test9", main(int argc, char* argv[]) test1(); test2(); test3(); test4(); test5(); test6(); test7(); test8(); test9(); 0; code></=></iostream></string>
48. 面试题48：最长不含重复字符的子字符串
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题48：最长不含重复字符的子字符串
// 题目：请从字符串中找出一个最长的不包含重复字符的子字符串，计算该最长子
// 字符串的长度。假设字符串中只包含从'a'到'z'的字符。
#include <string>
#include <iostream>
// 方法一：蛮力法
bool hasDuplication(const std::string& str, int position[]);
int longestSubstringWithoutDuplication_1(const std::string& str)
{
int longest = 0;
int* position = new int[26];
for(int start = 0; start < str.length(); ++start)
{
for(int end = start; end < str.length(); ++end)
{
int count = end - start + 1;
const std::string& substring = str.substr(start, count);
if(!hasDuplication(substring, position))
{
if(count > longest)
longest = count;
}
else
break;
}
}
delete[] position;
return longest;
}
bool hasDuplication(const std::string& str, int position[])
{
for(int i = 0; i < 26; ++i)
position[i] = -1;
for(int i = 0; i < str.length(); ++i)
{
int indexInPosition = str[i] - 'a';
if(position[indexInPosition] >= 0)
return true;
position[indexInPosition] = indexInPosition;
}
return false;
}
// 方法一：动态规划
int longestSubstringWithoutDuplication_2(const std::string& str)
{
int curLength = 0;
int maxLength = 0;
int* position = new int[26];
for(int i = 0; i < 26; ++i)
position[i] = -1;
for(int i = 0; i < str.length(); ++i)
{
int prevIndex = position[str[i] - 'a'];
if(prevIndex < 0 || i - prevIndex > curLength)
++curLength;
else
{
if(curLength > maxLength)
maxLength = curLength;
curLength = i - prevIndex;
}
position[str[i] - 'a'] = i;
}
if(curLength > maxLength)
maxLength = curLength;
delete[] position;
return maxLength;
}
// ====================测试代码====================
void testSolution1(const std::string& input, int expected)
{
int output = longestSubstringWithoutDuplication_1(input);
if(output == expected)
std::cout << "Solution 1 passed, with input: " << input << std::endl;
else
std::cout << "Solution 1 FAILED, with input: " << input << std::endl;
}
void testSolution2(const std::string& input, int expected)
{
int output = longestSubstringWithoutDuplication_2(input);
if(output == expected)
std::cout << "Solution 2 passed, with input: " << input << std::endl;
else
std::cout << "Solution 2 FAILED, with input: " << input << std::endl;
}
void test(const std::string& input, int expected)
{
testSolution1(input, expected);
testSolution2(input, expected);
}
void test1()
{
const std::string input = "abcacfrar";
int expected = 4;
test(input, expected);
}
void test2()
{
const std::string input = "acfrarabc";
int expected = 4;
test(input, expected);
}
void test3()
{
const std::string input = "arabcacfr";
int expected = 4;
test(input, expected);
}
void test4()
{
const std::string input = "aaaa";
int expected = 1;
test(input, expected);
}
void test5()
{
const std::string input = "abcdefg";
int expected = 7;
test(input, expected);
}
void test6()
{
const std::string input = "aaabbbccc";
int expected = 2;
test(input, expected);
}
void test7()
{
const std::string input = "abcdcba";
int expected = 4;
test(input, expected);
}
void test8()
{
const std::string input = "abcdaef";
int expected = 6;
test(input, expected);
}
void test9()
{
const std::string input = "a";
int expected = 1;
test(input, expected);
}
void test10()
{
const std::string input = "";
int expected = 0;
test(input, expected);
}
int main(int argc, char* argv[])
{
test1();
test2();
test3();
test4();
test5();
test6();
test7();
test8();
test9();
test10();
return 0;
}
</iostream></string>
50. 面试题50：字符串中第一个只出现一次的字符
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题50（一）：字符串中第一个只出现一次的字符
// 题目：在字符串中找出第一个只出现一次的字符。如输入"abaccdeff"，则输出
// 'b'。
#include <cstdio>
#include <string>
char FirstNotRepeatingChar(const char* pString)
{
if(pString == nullptr)
return '\0';
const int tableSize = 256;
unsigned int hashTable[tableSize];
for(unsigned int i = 0; i < tableSize; ++i)
hashTable[i] = 0;
const char* pHashKey = pString;
while(*(pHashKey) != '\0')
hashTable[*(pHashKey++)] ++;
pHashKey = pString;
while(*pHashKey != '\0')
{
if(hashTable[*pHashKey] == 1)
return *pHashKey;
pHashKey++;
}
return '\0';
}
// ====================测试代码====================
void Test(const char* pString, char expected)
{
if(FirstNotRepeatingChar(pString) == expected)
printf("Test passed.\n");
else
printf("Test failed.\n");
}
int main(int argc, char* argv[])
{
// 常规输入测试，存在只出现一次的字符
Test("google", 'l');
// 常规输入测试，不存在只出现一次的字符
Test("aabccdbd", '\0');
// 常规输入测试，所有字符都只出现一次
Test("abcdefg", 'a');
// 鲁棒性测试，输入nullptr
Test(nullptr, '\0');
return 0;
}
</string></cstdio>
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题50（二）：字符流中第一个只出现一次的字符
// 题目：请实现一个函数用来找出字符流中第一个只出现一次的字符。例如，当从
// 字符流中只读出前两个字符"go"时，第一个只出现一次的字符是'g'。当从该字
// 符流中读出前六个字符"google"时，第一个只出现一次的字符是'l'。
#include <cstdio>
#include <vector>
#include <limits>
using namespace std;
class CharStatistics
{
public:
CharStatistics() : index(0)
{
for(int i = 0; i < 256; ++i)
occurrence[i] = -1;
}
void Insert(char ch)
{
if(occurrence[ch] == -1)
occurrence[ch] = index;
else if(occurrence[ch] >= 0)
occurrence[ch] = -2;
index++;
}
char FirstAppearingOnce()
{
char ch = '\0';
int minIndex = numeric_limits<int>::max();
for(int i = 0; i < 256; ++i)
{
if(occurrence[i] >= 0 && occurrence[i] < minIndex)
{
ch = (char) i;
minIndex = occurrence[i];
}
}
return ch;
}
private:
// occurrence[i]: A character with ASCII value i;
// occurrence[i] = -1: The character has not found;
// occurrence[i] = -2: The character has been found for mutlple times
// occurrence[i] >= 0: The character has been found only once
int occurrence[256];
int index;
};
// ====================测试代码====================
void Test(const char* testName, CharStatistics chars, char expected)
{
if(testName != nullptr)
printf("%s begins: ", testName);
if(chars.FirstAppearingOnce() == expected)
printf("Passed.\n");
else
printf("FAILED.\n");
}
int main(int argc, char* argv[])
{
CharStatistics chars;
Test("Test1", chars, '\0');
chars.Insert('g');
Test("Test2", chars, 'g');
chars.Insert('o');
Test("Test3", chars, 'g');
chars.Insert('o');
Test("Test4", chars, 'g');
chars.Insert('g');
Test("Test5", chars, '\0');
chars.Insert('l');
Test("Test6", chars, 'l');
chars.Insert('e');
Test("Test7", chars, 'l');
return 0;
}
</int></limits></vector></cstdio>
67. 面试题67：把字符串转换成整数
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题67：把字符串转换成整数
// 题目：请你写一个函数StrToInt，实现把字符串转换成整数这个功能。当然，不
// 能使用atoi或者其他类似的库函数。
#include <cstdio>
long long StrToIntCore(const char* str, bool minus);
enum Status {kValid = 0, kInvalid};
int g_nStatus = kValid;
int StrToInt(const char* str)
{
g_nStatus = kInvalid;
long long num = 0;
if(str != nullptr && *str != '\0')
{
bool minus = false;
if(*str == '+')
str ++;
else if(*str == '-')
{
str ++;
minus = true;
}
if(*str != '\0')
num = StrToIntCore(str, minus);
}
return (int)num;
}
long long StrToIntCore(const char* digit, bool minus)
{
long long num = 0;
while(*digit != '\0')
{
if(*digit >= '0' && *digit <= 10 '9') { int flag="minus" ? -1 : 1; num="num" * + (*digit - '0'); if((!minus &&> 0x7FFFFFFF)
|| (minus && num < (signed int)0x80000000))
{
num = 0;
break;
}
digit++;
}
else
{
num = 0;
break;
}
}
if(*digit == '\0')
g_nStatus = kValid;
return num;
}
// ====================测试代码====================
void Test(const char* string)
{
int result = StrToInt(string);
if(result == 0 && g_nStatus == kInvalid)
printf("the input %s is invalid.\n", string);
else
printf("number for %s is: %d.\n", string, result);
}
int main(int argc, char* argv[])
{
Test(nullptr);
Test("");
Test("123");
Test("+123");
Test("-123");
Test("1a33");
Test("+0");
Test("-0");
//有效的最大正整数, 0x7FFFFFFF
Test("+2147483647");
Test("-2147483647");
Test("+2147483648");
//有效的最小负整数, 0x80000000
Test("-2147483648");
Test("+2147483649");
Test("-2147483649");
Test("+");
Test("-");
return 0;
}
</=></cstdio>
Original: https://www.cnblogs.com/agui125/p/12024893.html
Author: 风御之举
Title: 字符串
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