JZ-C-41

剑指offer第四十一题：和为s的两个数字：输入一个递增排序的数组和一个数字，在数组中查找两个数，使得它们的和正好等于s。若有多对，输出任意一对即可

//============================================================================
// Name        : JZ-C-41.cpp
// Author      : Laughing_Lz
// Version     :
// Copyright   : All Right Reserved
// Description :和为s的两个数字：输入一个递增排序的数组和一个数字，在数组中查找两个数，使得它们的和正好等于s。若有多对，输出任意一对即可
//============================================================================

#include <iostream>
#include <stdio.h>
using namespace std;

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

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

    while (ahead > behind) {
        long long curSum = data[ahead] + data[behind];

        if (curSum == sum) {
            *num1 = data[behind];
            *num2 = data[ahead];
            found = true;
            break;
        } else if (curSum > sum)
            ahead--;
        else
            behind++;
    }

    return found;
}

// ====================测试代码====================
void Test(char* testName, int data[], int length, int sum,
        bool expectedReturn) {
    if (testName != NULL)
        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. 
");
            else
                printf("Failed. 
");
        } else
            printf("Passed. 
");
    } else
        printf("Failed. 
");
}

// 存在和为s的两个数字，这两个数字位于数组的中间
void Test1() {
    int data[] = { 1, 2, 4, 7, 11, 15 };
    Test("Test1", data, sizeof(data) / sizeof(int), 15, true);
}

// 存在和为s的两个数字，这两个数字位于数组的两段
void Test2() {
    int data[] = { 1, 2, 4, 7, 11, 16 };
    Test("Test2", data, sizeof(data) / sizeof(int), 17, true);
}

// 不存在和为s的两个数字
void Test3() {
    int data[] = { 1, 2, 4, 7, 11, 16 };
    Test("Test3", data, sizeof(data) / sizeof(int), 10, false);
}

// 鲁棒性测试
void Test4() {
    Test("Test4", NULL, 0, 0, false);
}

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

    return 0;
}