【C++】string的使用

1、string的使用

#define _CRT_SECURE_NO_WARNINGS 1
#include <iostream>
#include<string>
using namespace std;void Test1()
{string s1;string s2("hello");cin >> s1;cout << s1 << endl;//strcat【字符串拼接】string ret1 = s1 + s1;cout << ret1 << endl;string ret2 = s1 + "我来了";cout << ret2 << endl;
}void Test2()
{string s1("Hello String");string s2 = "Hello String";string s3("Hello String");//遍历string
//方法1：【数组】//读for (int i = 0; i < s1.size(); i++){cout << s1[i];}cout << endl;//写for (int i = 0; i < s1.size(); i++){s1[i]++;}cout << s1 << endl;cout << "------------" << endl;//方法2：【迭代器】//读string::iterator it = s2.begin();while (it != s2.end()){cout << *it;++it;}cout << endl;//写it = s2.begin();while (it != s2.end()){*it = 'A';++it;}cout << s2 << endl;cout << "------------" << endl;//方法3：【范围for】//读for (auto ch : s3){cout << ch;}cout << endl;//写for (auto& ch : s3){++ch;}cout << s3 << endl;
}void Test3()
{string s1("Hello String");//反向迭代器//string::reverse_iterator rit = s1.rbegin();auto rit = s1.rbegin();while (rit != s1.rend()){cout << *rit;++rit;}cout << endl;
}void func(const string& s)
{//const迭代器//string::const_iterator it = s.begin();auto it = s.begin();while (it != s.end()){cout << *it;++it;}cout << endl;//const反向迭代器//string::const_reverse_iterator rit = s.rbegin();auto rit = s.rbegin();while (rit != s.rend()){cout << *rit;++rit;}cout << endl;
}void Test4()
{string s1("Hello String");func(s1);
}void Test5()
{string s1("Hello String");string s2(s1);string s3(s1, 6, 6); //从下标为6的地方起始，往后输出3个字符cout << s3 << endl;string s4(s2, 6, 3);cout << s4 << endl;string s5("Hello String XXXXXXXXXXXXXXXXXXXXXXXXX");//strlen不能用于自定义类型cout << string(s5, 6, s5.size() - 1) << endl; //从起始位置往后打印完cout << string(s5, 6) << endl; //不给输出多少，默认从起始位置全部输出完string s6(10, 'A'); //n个字符cout << s6 << endl;cout << string("Hello String", 5) << endl; //从字符串中取前n个string s8(s1.begin(), s1.end()); //用迭代器拷贝cout << s8 << endl; string s9(++s1.begin(), --s1.end()); //迭代器可以控制范围cout << s9 << endl;//其他赋值s9 = s8;s9 = "XXX";s9 = 'Y';
}void Test6()
{string s1("Hello String");cout << s1.size() << endl; //12【不算\0】cout << s1.length() << endl; //12cout << s1.capacity() << endl; //15s1.clear(); //只清数据，不清空间s1 = "LMY";cout << s1.capacity() << endl; //15/*string filename;cin >> filename;FILE* fout = fopen(filename.c_str(), "r");*/
}void Test7()
{//string的扩容string s;s.reserve(100); //【保留空间，提前开空间】size_t old = s.capacity();cout << "初始容量：" << s.capacity() << endl;for (size_t i = 0; i < 100; i++){s.push_back('x');if (s.capacity() != old){cout << "扩容成功：" << s.capacity() << endl;old = s.capacity();}}s.reserve(10); //VS不能缩空间cout << "缩后空间容量：" << s.capacity() << endl;
}void Test8()
{string s("Hello World");cout << "resize前：" << s.size() << endl; //11cout << "resize前：" << s.capacity() << endl << endl; //15//size < resize < capacity//s.resize(13); //【不给值，s里默认存\0】s.resize(13, 'x');cout << s << endl;cout << "resize后：" << s.size() << endl; //13cout << "resize后：" << s.capacity() << endl << endl; //15//capacity < resize【相当于插入数据】s.resize(20, 'x');cout << s << endl;cout << "resize后：" << s.size() << endl; //20cout << "resize后：" << s.capacity() << endl << endl; //31//resize < size【相当于删除数据】s.resize(5);cout << s << endl;cout << "resize后：" << s.size() << endl; //5cout << "resize后：" << s.capacity() << endl << endl; //31
}void Test9()
{//insert/erase/repalce能不用就尽量不用，因为他们都涉及挪动数据，效率不高string s("Hello World");s.insert(0, 1, 'L');s.insert(s.begin(), 'L');cout << s << endl;s.erase(8,2); //【从下标为8的位置开始，往后删俩】cout << s << endl;s.erase(5); //【不给删除个数，默认从起始位置开始往后全删】cout << s << endl << endl;//空格替换为20%string s2("The quick brown fox jumps over a lazy dog.");string s3;for (auto ch : s2){if (ch != ' '){s3 += ch;}else{s3 += "20%";}}//swap(s2, s3);//s2 = s3;//s2.assign(s3);printf("s2：%p\n", s2.c_str());printf("s3：%p\n", s3.c_str());s2.swap(s3); //交换的是指针cout << s2 << endl;printf("s2：%p\n", s2.c_str());printf("s3：%p\n", s3.c_str());
}void Test10()
{string s1("test.cpp.tar.zip");size_t i = s1.find('.');string s2 = s1.substr(i);cout << s2 << endl;size_t j = s1.rfind('.');string s3 = s1.substr(j);cout << s3 << endl << endl;//分离协议、域名、资源名string s("https://cplusplus.com/reference/string/string/");string sub1, sub2, sub3;size_t i1 = s.find(':');if (i1 != string::npos)sub1 = s.substr(0, i1);elsecout << "没有找到i1" << endl;size_t i2 = s.find('/', i1 + 3);if (i2 != string::npos)sub2 = s.substr(i1 + 3, i2 - (i1 + 3)); //【左闭右开区间，一减就是个数】elsecout << "没有找到i2" << endl;sub3 = s.substr(i2 + 1);cout << sub1 << endl;cout << sub2 << endl;cout << sub3 << endl;
}void Test11()
{string str;//getline(cin, str); //【读取一行】getline(cin, str, '!'); //【读取到'!'结束】cout << str;
}int main()
{Test11();return 0;
}

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2、string的OJ题

（1）字符串中的第一个唯一字符

class Solution
{
public:int firstUniqChar(string s){int count[26] = { 0 };//统计次数/*for (int i = 0; i < s.size(); i++){count[s[i] - 'a']++;}*/for (auto ch : s){count[ch - 'a']++;}for (int i = 0; i < s.size(); i++){if (count[s[i] - 'a'] == 1){return i;}}return -1;}
};

（2）仅仅反转字母

class Solution
{
public:bool IsLater(char ch){if (ch >= 'a' && ch <= 'z')return true;if (ch >= 'A' && ch <= 'Z')return true;return false;}string reverseOnlyLetters(string s){if (s.empty())return s;int begin = 0;int end = s.size() - 1;while (begin < end){while (begin < end && !IsLater(s[begin])){begin++;}while (begin < end && !IsLater(s[end])){end--;}swap(s[begin], s[end]);begin++;end--;}return s;}
};

（3）字符串最后一个单词的长度

• scanf / cin 遇到空格和换行符自动停止读取，空格和换行符后边的内容会被放在缓冲区中不被此次读取。但是 getline 可以。
• getline(cin, s);【可以读取一整行，遇到空格不停，遇到换行符才停止读取内容】
• getline(cin, str, '!'); 【读取到’!'结束】

#include <iostream>
#include <string>
using namespace std;int main()
{string s;getline(cin, s);size_t i = s.rfind(' ');if (i != string::npos){cout << s.size() - (i + 1) << endl;}else{cout << s.size() << endl;}return 0;
}

（4）字符串相加

class Solution
{
public:string addStrings(string num1, string num2){string str;int next = 0; //进位int end1 = num1.size() - 1;int end2 = num2.size() - 1;while (end1 >= 0 || end2 >= 0){int x1 = end1 >= 0 ? num1[end1] - '0' : 0; //转成整数int x2 = end2 >= 0 ? num2[end2] - '0' : 0; //转成整数int ret = x1 + x2 + next;//进位next = ret / 10;ret = ret % 10;//头插//str.insert(0,1,ret+'0'); //转成字符//尾插str += ret + '0';end1--;end2--;}if (next == 1){//str.insert(0,1,next+'0');str += next + '0';}reverse(str.begin(), str.end()); //逆置字符串return str;}
};

（5）验证回文串

class Solution
{
public://判断是不是字母bool IsLatter(char ch){return (ch >= 'A' && ch <= 'Z') || (ch >= 'a' && ch <= 'z') || (ch >= '0' && ch <= '9');}bool isPalindrome(string s){//大写字母全部转小写for (auto& ch : s) //要加引用才能修改{if (ch >= 'A' && ch <= 'Z'){ch += 32;}}int begin = 0;int end = s.size() - 1;while (begin < end){while (begin < end && !IsLatter(s[begin])){begin++;}while (begin < end && !IsLatter(s[end])){end--;}if (s[begin] == s[end]){begin++;end--;}else{return false;}}return true;}
};

（6）字符串转整型数字

class Solution
{
public:int StrToInt(string str){int len = str.size();int flag = 1; //区分正负数int sum = 0;if (str[0] == '-'){flag = -1;}//从个位开始for (int i = len - 1; i >= 0; i--){if (str[i] == '+' || str[i] == '-'){continue;}if (str[i] >= '0' && str[i] <= '9'){//判断位数int n = 1;for (int j = len - 1; j > i; j--){n = n * 10;}sum += (str[i] - '0') * n;}else{return 0;}}return flag * sum;}
};

（7）反转字符串Ⅰ

class Solution
{
public:void reverseString(vector<char>& s){int begin = 0;int end = s.size() - 1;while (begin < end){swap(s[begin], s[end]);begin++;end--;}}
};

（8）反转字符串Ⅱ

class Solution
{
public://翻转start到end区间的字符串void Reverse(string& s, int start, int end){while (start < end){swap(s[start], s[end]);start++;end--;}}string reverseStr(string s, int k){int len = s.size();for (int i = 0; i < len; i += 2 * k){if (i + k < len)Reverse(s, i, i + k - 1);elseReverse(s, i, len - 1);}return s;}
};

（9）反转字符串中的单词

class Solution
{
public:void reverse(string& s, int begin, int end){while (begin < end){swap(s[begin], s[end]);begin++;end--;}}string reverseWords(string s){int begin = 0;int end = 0;while (end < s.size()){//找空格end = s.find(' ', begin);if (end == string::npos){end = s.size();break;}//反转reverse(s, begin, end - 1);begin = end + 1;}//反转最后一个单词reverse(s, begin, end - 1);return s;}
};

（10）字符串相乘

• 两个数相乘之后的结果的位数肯定不会超过两个数位数之和，定义一个长度为两数位数之和的字符串，然后从尾部开始迭代，通过取余和整除的操作依次计算各位结果，最后通过 substr 从不为零的位置返回字符串。

class Solution
{
public:string multiply(string num1, string num2){//两个数相乘之后的结果的位数肯定不会超过两个数位数之和//定义一个长度为两数位数之和的字符串，然后从尾部开始迭代，通过取余和整除的操作依次计算各位结果//最后通过substr从不为零的位置返回字符串int len1 = num1.size();int len2 = num2.size();string arr(len1 + len2, '0');//从后按位迭代for (int i = len1 - 1; i >= 0; i--){for (int j = len2 - 1; j >= 0; j--){int tmp = (num1[i] - '0') * (num2[j] - '0') + (arr[i + j + 1] - '0');arr[i + j + 1] = tmp % 10 + '0';arr[i + j] = tmp / 10 + (arr[i + j] - '0') + '0';}}//取字符串for (int i = 0; i < len1 + len2; i++){if (arr[i] != '0'){return arr.substr(i);}}return "0";}
};