C++ 数组排序 - 将“bbba”和“0001”视为不正确排序的问题

Question

 //is the array ordered?
    //if there are one or zero elements, inherently sorted
    //can't compare to other elements if they don't exist
    if ( len == 0 || len ==1 )
    {
        return true;
        cout<<"isOrdered"<<endl;
    }
    //if more than 1 element <> can do a comparison
    for (int i=0; i<len; i++)
    {
          if ( sArr[i] > sArr[i+1])
          {
            return false;
            cout<<"notOrdered"<<endl;
          }
          else if (sArr[i]<=sArr[i+1])
            {
              return true; //here it must be ordered.
            }
    }

}

sort logic taken from: https://www.geeksforgeeks.org/program-check-array-sorted-not-iterative-recursive/

This code is in a larger function, which turns strings and unsigned ints into arrays, and checks if the order of elements is ASCII ordered from lowest to highest.

Problem is that it does not treat "abba" or "111110" as not ordered cases. it continually returns true for cases like this and I don't understand why. Doesn't the code compare element by element to the one right next to it?

Answer 1

You don't need to test for 0 or 1 at start (since the for loop already has the i<len condition and will be false for both 0 and 1) but you do need to move the return true out of the loop. You don't know that it's sorted until you've compared all elements:

for(int i=1; i<len; ++i) {
    if(sArr[i-1] > sArr[i]) return false;
}
return true;

Answer 2

from your code, you need to return at least there is one out of order and then need to return true when all are okay.

** also need to run only till (length -1)

for (int i=0; i<len-1; i++)
{
      if ( sArr[i] > sArr[i+1])
      {
        cout<<"notOrdered"<<endl;
        return false;
      }
}
cout<<"isOrdered"<<endl;
return true;

Answer 3

Testing for 0 and 1 is a good idea in the recursive case. This function could be called recursively, so you can leave that out. You have left out the recursion and used a for loop instead. Here is the corrected code, which waits until after the for loop to return true if the current pairs are sorted correctly.

#include <iostream>

using namespace std;

bool IsArraySorted(int sArr[], int len)
{
  //is the array ordered?

  //if more than 1 element <> can do a comparison
  for (int i = 0; i < len - 1; i++) {
    if (sArr[i] > sArr[i + 1]) {
      cout << "notOrdered" << endl; // call before return :-)

      return false;
    }
  }
  cout << "Ordered" << endl;

  return true; //here it must be ordered.

}