所以问题是我正在提取一个样本文件，要求输入一个用户帐号。现在，输出是：
What is the filename of account numbers? sample.txt
What is the account number you are looking for? 5552012
-858993460
4520125
5658845
7895122
8451277
1302850
8080152
4562555
5552012
5050552
7825877
1250255
1005231
6545231
3852085
7576651
7881200
4581002
The account number you were looking for is: 1
Press any key to continue . . .
下面输出的数字：“帐号是什么？”由于'for'循环用于调试目的。输出的问题是顶部数字（-858993460）不是sample.txt文件中存在的数字。应该在那里的数字是5658845。我猜测负数是最小的可能。
这部分是我正在使用的代码。似乎冒泡排序算法不能正常工作。但是，按照指示，我们将遵循我所做的书籍示例。我已经多次检查算法函数并发现没有错误，但我的评估可能出错了。此问题导致更大的问题，导致无法在已排序的数组中找到正确的帐号，该数组在当前状态下返回1，这不是存在的数字。
#include <stdafx.h>
#include <stdio.h>
#include <fstream>
#include <string>
#include <algorithm>
#include <iostream>
using namespace std;
//prototypes
void sortAcctNums(int accounts[], int arraySize);
bool findAcctNum(int accounts[], int numElems, int accountNumSearched);
//main part of the program. This is where all the magic happens
int main(){
string fileName;
int accountNumSearched, count = 0;
const int arraySize = 18;
int accounts[arraySize];
int location;
fstream inputFile(fileName, ios::in);
cout << "What is the filename of account numbers? ";
cin >> fileName;
inputFile.open(fileName);
if (inputFile.is_open()){//makes sure the file is able to be read, if not then requests user to try again
cout << "What is the account number you are looking for? ";
cin >> accountNumSearched;
while (inputFile >> accounts[count]){
count++;//populates the array
}
inputFile.close();//closes the file stream
sortAcctNums(accounts, arraySize);//sorts the account numbers
location = findAcctNum(accounts, arraySize, accountNumSearched);//assigns the value of the fundAcctNum function to location
if (location == -1){
cout << "The account number could not be found" << endl;
exit;
}
else
cout << "The account number you were looking for is: " << location << endl;
}
else
cout << "Error opening file. Please try again. " << endl;
return 0;
}
//this function sorts the account numbers with a bubble sort algorithm
void sortAcctNums(int accounts[], int arraySize){
bool swap;
int temp;
do{
swap = false;
for (int i = 0; i < (arraySize - 1); i++){
if (accounts[i] > accounts[arraySize + 1]){
temp = accounts[i];
accounts[i] = accounts[arraySize + 1];
accounts[arraySize + 1] = temp;
swap = true;
}
}
} while (swap);
//this loop is only for testing purposes and should be removed during final build
for (int i = 0; i < arraySize; i++){
cout << accounts[i] << endl;
}
}
//This function searches the sorted array for the specified account number with a Binary sort algorithm
bool findAcctNum(int accounts[], int numElems, int accountNumSearched){
int first = 0,
last = numElems - 1,
middle,
position = -1;
bool found = false;
while (!found && first <= last){
middle = (first + last) / 2;
if (accounts[middle] == accountNumSearched){
found = true;
position = middle;
}
else if (accounts[middle] > accountNumSearched)
last = middle - 1;
else
first = middle + 1;
}
return position;
}//end search
答案 0 :(得分：2)
您的冒泡排序完全没有问题。你有：
do{
swap = false;
for (int i = 0; i < (arraySize - 1); i++){
if (accounts[i] > accounts[arraySize + 1]){
temp = accounts[i];
accounts[i] = accounts[arraySize + 1];
accounts[arraySize + 1] = temp;
swap = true;
}
}
} while (swap);
这不是泡泡排序。您需要嵌套的for循环才能进行冒泡排序。回到你的例子，确保你完全实现它。典型的冒泡排序看起来与此相似：
swap = true;
for (int i = 0; i < (arraySize - 1) && swap; ++i)
{
swap = false;
for (int j = 0; j < (arraySize - i - 1); ++j)
{
if (array[j] > array[j+1])
{
temp = array[j];
array[j] = array[j+1];
array[j+1] = temp;
swap = true;
}
}
}
您在阵列中获取该虚假号码的原因是您将当前项目（accounts[i]）与accounts[arraySize+1]进行比较。在这种情况下，您需要与accounts[20]进行比较。由于阵列中只有18个项目，因此您需要与阵列末尾存储在内存中的某些随机值进行比较。负数不是最小的int。
答案 1 :(得分：1)
正确的计划是：
#include <stdafx.h>
#include <stdio.h>
#include <fstream>
#include <string>
#include <algorithm>
#include <iostream>
using namespace std;
//prototypes
void sortAcctNums(int accounts[], size_t count);
int findAcctNum(int accounts[], int numElems, int accountNumSearched);
//main part of the program. This is where all the magic happens
int main(){
string fileName;
int accountNumSearched, count = 0;
const int arraySize = 18;
int accounts[arraySize];
int location;
fstream inputFile(fileName, ios::in);
cout << "What is the filename of account numbers? ";
cin >> fileName;
inputFile.open(fileName);
if (inputFile.is_open()){//makes sure the file is able to be read, if not then requests user to try again
cout << "What is the account number you are looking for? ";
cin >> accountNumSearched;
while (inputFile >> accounts[count]){
count++;//populates the array
}
inputFile.close();//closes the file stream
sortAcctNums(accounts, sizeof(accounts)/sizeof(*accounts));//sorts the account numbers
location = findAcctNum(accounts, count, accountNumSearched);//assigns the value of the fundAcctNum function to location
if (location == -1){
cout << "The account number could not be found" << endl;
exit;
}
else
cout << "The account number you were looking for is: " << accounts[location] << endl
}
else
cout << "Error opening file. Please try again. " << endl;
return 0;
}
//this function sorts the account numbers with a bubble sort algorithm
void sortAcctNums(int accounts[], size_t count){
bool swap = true;
int temp;
while (count-- && swap){
swap = false;
for (size_t i = 0; i < count; ++i){
if (accounts[i] > accounts[i + 1]){
std::iter_swap(accounts + i, accounts + i + 1);
swap = true;
}
}
}
}
//This function searches the sorted array for the specified account number with a Binary search algorithm
int findAcctNum(int accounts[], int numElems, int accountNumSearched){
int first = 0,
last = numElems - 1,
middle,
position = -1;
bool found = false;
while (!found && first <= last){
middle = (first + last) / 2;
if (accounts[middle] == accountNumSearched){
found = true;
position = middle;
}
else if (accounts[middle] > accountNumSearched)
last = middle - 1;
else
first = middle + 1;
}
return position;
}//end search
答案 2 :(得分：0)
冒泡排序应如下所示：
for (int i = 0 ; i < (arraySize - 1); i++) {
for (int j = 0 ; j < arraySize - i - 1; j++) {
if (array[j] > array[j+1]) {
int swap = array[j];
array[j] = array[j+1];
array[j+1] = swap;
}
}
}
尝试一下，看看你是否仍然会收到不好的结果。