"Prove safety offer" fifty-seventh title I: and the two numbers s

// face questions 57 (a): s, and two numbers
 // Title: Enter a sorted array and incrementing a number s, find the two numbers in the array, such that they
 // and exactly s. If a plurality of numbers and equal to s, the output can be any pair. 

#include <cstdio>

bool FindNumbersWithSum(int data[], int length, int sum,
                        int* num1, int* num2)
{
    bool found = false;
    if (length < 1 || num1 == nullptr || num2 == nullptr)
        return found;

    int ahead = 0;
    int behind = length - 1;

    while (ahead <= behind)
    {
        long long result = data[ahead] + data[behind];
        if (result == sum)
        {
            *num1 = data[ahead];
            *num2 = data[behind];
            found = true;
            break;
        }
        else if (result > sum)
            --behind;
        else
            ++ahead;
    }
    return found;
}

// ==================== test code ==================== 
void the Test ( const  char * TestName, int Data [], int length, int SUM, BOOL expectedReturn)
{
    if (testName != nullptr)
        printf("%s begins: ", testName);

    int num1, num2;
    int result = FindNumbersWithSum(data, length, sum, &num1, &num2);
    if (result == expectedReturn)
    {
        if (result)
        {
            if (num1 + num2 == sum)
                printf("Passed. \n");
            else
                printf("FAILED. \n");
        }
        else
            printf("Passed. \n");
    }
    else
        printf("FAILED. \n");
}

// present and s is the number two, the two numbers in the middle of the array 
void Test1 ()
{
    int data[] = { 1, 2, 4, 7, 11, 15 };
    Test("Test1", data, sizeof(data) / sizeof(int), 15, true);
}

// present and s is the number two, these two figures located two arrays 
void Test2 ()
{
    int data[] = { 1, 2, 4, 7, 11, 16 };
    Test("Test2", data, sizeof(data) / sizeof(int), 17, true);
}

// absent and s is the number of two 
void the Test3 ()
{
    int data[] = { 1, 2, 4, 7, 11, 16 };
    Test("Test3", data, sizeof(data) / sizeof(int), 10, false);
}

// robustness tests 
void Test4 ()
{
    Test("Test4", nullptr, 0, 0, false);
}

int main(int argc, char* argv[])
{
    Test1();
    Test2 ();
    Test3 ();
    Test4 ();

    return 0;
}

Test code

Analysis: The book ahead and behind written backwards.

The first set of numbers encountered, i.e. the maximum difference between the outermost two numbers, the product must be minimized.

class Solution {
public:
    vector<int> FindNumbersWithSum(vector<int> array,int sum) {
        
        int length = array.size();
        int ahead = 0;
        int behind = length - 1;
        vector<int> num;
        
        while (ahead < behind)
        {
            int curSum = array[ahead] + array[behind];
            if (curSum == sum)
            {
                num.push_back(array[ahead]);
                num.push_back(array[behind]);
                break;
            }
            else if (curSum > sum)
                --behind;
            else
                ++ahead;
        }
        return num;
    }
};

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