题解 | #参数解析#
参数解析
https://www.nowcoder.com/practice/668603dc307e4ef4bb07bcd0615ea677
对字符串的操作，无非就是substr了，找到start 与 end 位置 就ok了，难度偏简单吧，偏基础
#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
using namespace std;
string s1;
bool flag = true;
vector<string>ans;
int main() {
getline(cin, s1);
int start = 0;
int end = 0;
for (int i = 0; i < s1.size(); i++) {
if (flag && (s1[i] == ' '||i==s1.size()-1)) {
end = i;
if (s1[start] == '"' && s1[end - 1] == '"')
ans.push_back(s1.substr(start + 1, end - 1 - start - 1));
else if (i == s1.size() - 1)
ans.push_back(s1.substr(start, end - start + 1));
else
ans.push_back(s1.substr(start, end - start));
start = end + 1;
continue;
}
else if (flag && s1[i] != ' '&& s1[i] != '"') {
continue;
}
else if (s1[i] == '"') {
flag = !flag;
if(i==s1.size()-1)
ans.push_back(s1.substr(start+1, i - start-1));
continue;
}
else
continue;
}
int n = ans.size();
cout << n << endl;
for (auto c : ans)
{
cout << c << endl;
}
return 0;
}
投递字节跳动等公司9个岗位