出于某种原因,我无法正确地获得这种名称。谁能告诉我它有什么问题?据我所知,问题是字符串没有正确比较。我以前尝试过字符串比较,我知道这种代码应该可行。真的让我难过。
#include <iostream>
#include <fstream>
#include <string>
#include <vector>
using namespace std;
void sortNames(vector<string> &);
void main()
{
vector<string> namesList;
ifstream namesFile;
namesFile.open("Names.txt");
// Make sure the file exists.
if (namesFile)
{
// Get the names from the file.
string name;
while (getline(namesFile, name))
namesList.push_back(name);
// Sort the imported names.
sortNames(namesList);
for (int i = 0; i < namesList.size(); i++)
cout << namesList[i] << endl;
}
else
{
cout << "Data files are missing";
}
namesFile.close();
}
void sortNames(vector<string> &list)
{
for (int i = 0; i < list.size(); i++)
{
// Find the lowest value after i.
int lowIndex = i;
for (int j = i + 1; j < list.size(); j++)
{
string name = list[i];
string name2 = list[j];
if (name > name2)
lowIndex = j;
}
// Flip the elements if there was a value lower than i.
if (i != lowIndex)
{
string temp = list[i];
list[i] = list[lowIndex];
list[lowIndex] = temp;
}
}
}
答案 0 :(得分:5)
问题在于:这一行
string name = list[i];
应该是
string name = list[lowIndex];
您当前的实现将j
处的元素与您到目前为止找到的最小字符串进行比较,而不是将其与索引i
处的字符串进行比较。这是不正确的,因为它找不到最小的剩余字符串:相反,它找到vector
中的最后一个字符串,该字符串小于索引i
处的当前元素,这不是您想要的。
答案 1 :(得分:0)
而非string name = list[i];
,您需要string name = list[lowIndex];