1.简洁高效的方法(不过只能包含一个分隔符):
#include <vector>#include <string>#include <iostream>using namespace std;void SplitString(const string& s, vector<string>& v, const string& c){string::size_type pos1, pos2;pos2 = s.find(c);pos1 = 0;while(string::npos != pos2){v.push_back(s.substr(pos1, pos2-pos1));pos1 = pos2 + c.size();pos2 = s.find(c, pos1);}if(pos1 != s.length())v.push_back(s.substr(pos1));}int main(){string s = "a,b,c,d,e,f";vector<string> v;SplitString(s, v,","); //可按多个字符来分隔;for(vector<string>::size_type i = 0; i != v.size(); ++i)cout << v[i] << " ";cout << endl;//输出: a b c d e f}
当处理有空格的字符串时,还是很有用的!!
使用void SplitString(const string& s, vector& v, const string& c),和将v作为返回值都是可以的!
2.可包含多个分隔符的实现方式
#include <vector>#include <string>#include <iostream>using namespace std;vector<string> split(const string &s, const string &seperator){vector<string> result;typedef string::size_type string_size;string_size i = 0;while(i != s.size()){//找到字符串中首个不等于分隔符的字母;int flag = 0;while(i != s.size() && flag == 0){flag = 1;for(string_size x = 0; x < seperator.size(); ++x)if(s[i] == seperator[x]){++i;flag = 0;break;}}//找到又一个分隔符,将两个分隔符之间的字符串取出;flag = 0;string_size j = i;while(j != s.size() && flag == 0){for(string_size x = 0; x < seperator.size(); ++x)if(s[j] == seperator[x]){flag = 1;break;}if(flag == 0)++j;}if(i != j){result.push_back(s.substr(i, j-i));i = j;}}return result;}int main(){// string s = "a,b*c*d,e";string s;getline(cin,s);vector<string> v = split(s, ",*"); //可按多个字符来分隔;for(vector<string>::size_type i = 0; i != v.size(); ++i)cout << v[i] << " ";cout << endl;//输出: a b c d e}
3. 用C语言中的strtok 函数来进行分割
原型: char _strtok(char _str, const char *delim);strtok函数包含在头文件
#include <string.h>#include <stdio.h>int main(){char s[] = "a,b*c,d";const char *sep = ",*"; //可按多个字符来分割char *p;p = strtok(s, sep);while(p){printf("%s ", p);p = strtok(NULL, sep);}printf("\n");return 0;}//输出: a b c d
From https://www.cnblogs.com/carsonzhu/p/5859552.html
4 最优雅的实现
void split(const std::string& s, std::vector<std::string>& tokens, const std::string& delimiters = " "){std::string::size_type lastPos = s.find_first_not_of(delimiters, 0);std::string::size_type pos = s.find_first_of(delimiters, lastPos);while (std::string::npos != pos || std::string::npos != lastPos){tokens.push_back(s.substr(lastPos, pos - lastPos));lastPos = s.find_first_not_of(delimiters, pos);pos = s.find_first_of(delimiters, lastPos);}}
