输入一个递增排序的数组和一个数字s,在数组中查找两个数，使得它们的各正好是s.
如果有多对数字的和等于s,输出任意一对即可。
#include<iostream> #include<stdio.h> 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; } 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. \n"); else printf("Failed. \n"); } else printf("Passed. \n"); } else printf("Failed. \n"); } void Test1() { int data[] = {1, 2, 4, 7, 11, 15}; Test("Test1", data, sizeof(data) / sizeof(int), 15, true); } void Test2() { int data[] = {1, 2, 4, 7, 11, 16}; Test("Test2", data, sizeof(data) / sizeof(int), 17, true); } void Test3() { int data[] = {1, 2, 4, 7, 11, 16}; Test("Test3", data, sizeof(data) / sizeof(int), 10, false); } void Test4() { Test("Test4", NULL, 0, 0, false); } int main() { Test1(); Test2(); Test3(); Test4(); return 0; }
