C ++中的字符串比较如何工作？

Question

I am trying to solve this problem .

I am implementing it with strings. Here is my code snippet

string s,ss;
// s and ss both contains integer input.

while(s <= ss )
//while( s<=ss && s.size() <= ss.size())
{
    int i = inc, j = dec;   // inc and dec are middle values. both equal if odd else different

    while((s[j]-'0')==9 && i < len && j>=0){
        // for cases like 999
        s[i] = s[j] = '0';
        i++;
        j--;
    }
    if(j<0){
        s = "1" + s;
        int l = s[len-1] - '0';
        l++;
        //cout<<l<<"\n";
        s[len] = (l + '0');
    }
    else{
        int l = s[j] - '0';
        l++;
        s[i] = s[j] = (l+'0');
    }

    if(s <= ss)
        cout<<"out in wild "<<s<<" and "<<ss<<"\n";
}

cout<<s<<endl;

The problem that I am facing is when input is like 999 or 9999. The outer while loop keeps on looping even when the value of s increases, but if I add while( s<=ss && s.size() <= ss.size()) it works completely fine. Why is while(s<=ss) is not working? I rarely use the string class, so I don't understand it completely. Why don't string s= 101 and ss=99 stop the while loop?

Complete code link is here

Answer 1

You are comparing strings with lexicographical order , not numbers , so "101" is less than "99" (because '1' < '9') , eg

int main(){
    std::string s = "99";
    std::string ss = "101";

    std::cout << std::boolalpha << (s <= ss);  
}

Outputs false .

Notes:

• A better design for your program would be to manipulate numbers ( int or double ...) and not strings in the first place, so this kind of expressions would naturally work as you expect.

  Eg "101" + "99" is "10199", not "200" ...

• But if you really need strings, consider this post to sort strings containing numbers.

• As pointed by @Deduplicator, a program that needlessly overuses strings is sometimes called Stringly Typed

• Also see std::lexicographical_compare

---

Since your input explicitly only involves positive integers without leading 0 , writing a comparison function is trivial, something like : (untested)

/* Returns 1 if the integer represented by s1 > the integer represented by s2
*  Returns -1 if the integer represented by s1 < the integer represented by s2
*  Return 0 is both are equals
*
*  s1 and s2 must be strings representing positive integers without trailing 0
*/
int compare(const std::string& s1, const std::string& s2)
{
  if(s1.size() > s2.size())
      return 1;
  if(s2.size() > s1.size())
      return -1;

  for(std::size_t i = 0 ; i < s1.size() ; ++i)
  {
    if(s1[i] - '0' < s2[i] - '0')
      return 1;
    if(s2[i] - '0' < s1[i] - '0')
      return -1;
  }

  return 0;
}

Answer 2

While s and ss are string variables, they are compared character by character.

In the case that you mentioned being: s = "101" & ss = "99" , by first hand it will check the first character in each string, and as '1' < '9' it exit up with s < ss . I would advise you to convert those values to integers before comparison.

Answer 3

由于按字母顺序将s与ss进行比较，我建议您将尾部的一个字符与头部的一个字符进行比较（一个到一个地到达中间），以解决该问题。