thinking

First, reverse the whole string, and then reverse each word, mainly testing the hard coding ability.

Specific implementation code

#include <iostream>
#include <string>

using namespace std;
typedef long long LL;

void reverse(string &s) {
    int length = s.size();
    for (int i = 0; i < length / 2; i++) {
        char temp = s[i];
        s[i] = s[length - 1 - i];
        s[length - 1 - i] = temp;
    }
}

void reversed(string &s) {
    reverse(s);
    int length = s.size(), start = 0,end = 0;
    //cout << s << endl;
    bool flag = true;
    for (int i = 0; i < length; i++) {
        if (s[i] == ' ' && flag) {
            string temp = s.substr(start, i - start  );
            //cout << temp << endl;
            reverse(temp);
            s.replace(start, i - start, temp);
            flag = false;
        }
        else if (!flag) {
            start = i;
            flag = !flag;
        }
    }
    string temp = s.substr(start, length - start);
    //cout << temp << endl;
    reverse(temp);
    s.replace(start , length - start, temp);
}


int main() {
    ios::sync_with_stdio(false);
    string s;
    getline(cin, s);
    reversed(s);
    cout << s << endl;
    system("pause");
    return 0;
}

My implementation is a bit stupid, but I have practiced the use of string class. reverse can use libraries, and substr functions can also be used. The end and start variables are also redundant.
Here is the code written by the simplified version of the big guy!

Link:[https://www.nowcoder.com/questionTerminal/ee5de2e7c45a46a090c1ced2fdc62355](https://www.nowcoder.com/questionTerminal/ee5de2e7c45a46a090c1ced2fdc62355)
//Source: niuke.com
int main() {
    string str;
    while (getline(cin, str)) {
        reverse(str.begin(), str.end());
        int i = 0, j = i;
        while (i<str.size()) {
            while (i<str.size() && str[i] == ' ')
                ++i;
            j = i;
            while (i<str.size() && str[i] != ' ')
                ++i;
            reverse(str.begin() + j, str.begin() + i);
            j = i;
        }
        cout << str << endl;
    }
    return 0;
}