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  1. #include <iostream>
  2. #include <cstdlib>
  3. using namespace std;
  4.  
  5. const int A[] = {4, 2, 5, 3, 7, 4, 2, 6, 8, 4, 1, 10, 10, 10, 20, 11, 15, 13, 28, 99, 11 };
  6. const int B[] = {10, 99, 20, 30, 11, 100, 150, 101, 125};
  7. const int C[] = {10, 20, 30, 40, 5};
  8.  
  9. int find_pivot(const int a[], int n)
  10. {
  11.     int lo = 0, hi = n-1;
  12.     for (int i=1;i<n && lo<=hi; ++i)
  13.     {
  14.         // find the low-maximum that is no greater than our
  15.         //  current high-minimum.
  16.         if (a[i] > a[lo] && a[hi] >= a[i])
  17.             lo = i;
  18.  
  19.         // likewise, find the hi-minimum that is not less than
  20.         //  our current low-maximum.
  21.         if (a[(n-1)-i] <= a[hi] && a[lo] < a[(n-1)-i])
  22.             hi = (n-1)-i;
  23.     }
  24.     return (lo == hi || a[lo] == a[hi] ? hi : -1);
  25. }
  26.  
  27. int main(int argc, char* argv[])
  28. {
  29.     int n = find_pivot(A, sizeof(A)/sizeof(A[0]));
  30.     if (n >=0)
  31.         cout << "A[" << n << "] = " << A[n] << endl;
  32.     else
  33.         cout << "A has no viable pivot.";
  34.  
  35.     n = find_pivot(B, sizeof(B)/sizeof(B[0]));
  36.     if (n >=0)
  37.         cout << "B[" << n << "] = " << B[n] << endl;
  38.     else
  39.         cout << "B has no viable pivot.";
  40.  
  41.     n = find_pivot(C, sizeof(C)/sizeof(C[0]));
  42.     if (n >=0)
  43.         cout << "C[" << n << "] = " << C[n] << endl;
  44.     else
  45.         cout << "C has no viable pivot.";
  46.  
  47.     return EXIT_SUCCESS;
  48. }
Success #stdin #stdout 0s 3296KB
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stdin
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Standard input is empty
stdout
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A[11] = 10
B[5] = 100
C has no viable pivot.
https://ideone.com/akYXVp
language:
C++ (gcc 8.3)
created:
10 years ago
visibility:
secret

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