对自定义对象的向量进行排序

问题描述

如何对包含自定义(即用户定义)对象的向量进行排序.
可能应该使用标准 STL 算法排序以及谓词(函数或函数对象)，该谓词将对自定义对象中的一个字段(作为排序的键)进行操作.
我在正确的轨道上吗?

How does one go about sorting a vector containing custom (i.e. user defined) objects.
Probably, standard STL algorithm sort along with a predicate (a function or a function object) which would operate on one of the fields (as a key for sorting) in the custom object should be used.
Am I on the right track?

推荐答案

一个使用 的简单示例std::sort

struct MyStruct
{
    int key;
    std::string stringValue;

    MyStruct(int k, const std::string& s) : key(k), stringValue(s) {}
};

struct less_than_key
{
    inline bool operator() (const MyStruct& struct1, const MyStruct& struct2)
    {
        return (struct1.key < struct2.key);
    }
};

std::vector < MyStruct > vec;

vec.push_back(MyStruct(4, "test"));
vec.push_back(MyStruct(3, "a"));
vec.push_back(MyStruct(2, "is"));
vec.push_back(MyStruct(1, "this"));

std::sort(vec.begin(), vec.end(), less_than_key());

正如 Kirill V. Lyadvinsky 指出的，您可以为 MyStruct 实现 operator<，而不是提供排序谓词:

---

As Kirill V. Lyadvinsky pointed out, instead of supplying a sort predicate, you can implement the operator< for MyStruct:

struct MyStruct
{
    int key;
    std::string stringValue;

    MyStruct(int k, const std::string& s) : key(k), stringValue(s) {}

    bool operator < (const MyStruct& str) const
    {
        return (key < str.key);
    }
};

使用此方法意味着您可以简单地按如下方式对向量进行排序:

Using this method means you can simply sort the vector as follows:

std::sort(vec.begin(), vec.end());

Edit2:正如 Kappa 建议的，您还可以通过重载 > 运算符并稍微更改排序调用来按降序对向量进行排序:

As Kappa suggests you can also sort the vector in the descending order by overloading a > operator and changing call of sort a bit:

struct MyStruct
{
    int key;
    std::string stringValue;

    MyStruct(int k, const std::string& s) : key(k), stringValue(s) {}

    bool operator > (const MyStruct& str) const
    {
        return (key > str.key);
    }
};

你应该调用 sort 为:

And you should call sort as:

std::sort(vec.begin(), vec.end(),greater<MyStruct>());

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