算法面试题：一个List<Student>，要求删除里面的男生，不用Linq和Lamda，求各种解，并说明优缺点！
算法面试题：一个List，要求删除里面的男生，不用Linq和Lamda，求各种解，并说明优缺点！
解题思路
这是群里某位小伙伴去面试碰到的面试题，从题目本身来看，面试官应该是要考察面试者对泛型 List 的理解程度，也算是对基础的理解。这里面还是有很多需要考察的知识点，没关系，我们走一步看一步。
首先，题目没有限定数量，我们知道，只谈毒性不谈剂量的都是耍流氓，那我们就假定要处理的数据有 10000 条好了。
数据准备好了
static readonly string text = "1000100010101001101101011110111111100000100010001101010101001101111010100001010101100101011011101011111010101000111100110101001001000010100000010111010110011010111100000101110011011100100010110101100010111010100010110000001011110110000011110001101001000000111100110101101111111000011100010110010000000101111010111011100000101111000011101101110100111010101001010000001101000100111011111101010100011011000001001010100100001001011001100011001010111110110100101101101010010100010101011100110101110011011101001001001110101101011010101001000011001111010101110101010001111010110001011100010001000011000101000011010011011000100011111110111010001111111000100011110000100010010101000111110111111101111111000000001011010101000100100101010100000111101111010011011011100000001001000110011100110110001110100100110010011100000110111010010011100011001100010001011100010010001011100110101110000001000010001111101010110000111111110101110100111101110000111000110011110101001110010001010100110110010100011110110111010010011011100011010111011010100110011010001110011101000100010100011110001000101010100000010111011010101100001111011000001110110010011111011000010001101100001101100001010010000001001101110101110001100000100100100110101110110110101001111011011010111000100110011011111001010001100001011000110000100011000010010101011010100111001000110001011101010100010101111101100101001110100111000011110101111111001011100001001001100110011001010001111000110000101111010101010011101010111000000001001111001111101010101010101101100101010111001011100000010001011001000001110101000110111000011110101010110101101001001101001110000010000000010011111110101001111111000100100010001100101111011111000110100000000000110010011101010010011100000000111111000001011100000101100001100010101111010110000100011101101101111011100010011111100100100001001001011001000000010100110101110101101100001000011110001100010101011000100000000011101001101101100101010011000101100100001010100011100111110011100110110011100011100011110001101100000001111011100111110011111100001100001111101101111101110001111011010100011000101000111110110111011000011110010110010010100011100101111000101110010110100000011010000101110100001101001001110010110101101011110110001111101101000000011011010001111000010110000011110100101011011100100111001101001101110011101010001001001010010111100010111111111001101010000011000000000101111101011110111001001000010100000010100000110101001101100001111001011100000101010000010000111101100111011101001111000011110111011100011101110000101111000111011101010100000110011000010001101101110000111010111100001011110001101100001000110101010010001010000000001110111011110001110101000000101110010100110010110101011100000100010011100001001010100110100100101111010111100000101010100100101001010110000101110110010110100101101111000010100001010011111111010011101001001011001000101000010000000100111001011010110000010001110010011100010111111100101001010100111000100011001111100101100011100000101100110010101100110111100000001111111011000100010100110111011011001000010110100000001111011010001010101110010111101000010110001010110011001010100101010001110100110100111101010011011110011101011111100101011010101101010010000100110111000011011110110000010110110000001010101011110001110001101100110111011010001001010011110100011111110110110000000111100111100111001110111011011010010100101110001111001110001100110011111001001000110011101111111100111011001001010011101100010000111010111000111101010101100010100101000001000111000010001010110110100010100011010110111110111111101001101000010111000111111011001100001111110011111100000110100110000011010110100111001001100111001010011011011000111100001001111100100101101101010111101001101001000101000100101000010110010100001111001010010111101001100011100010111001110100111111101110011100000011110110100001100001011110110100011010010001100000011111110110001001101001010010010000001000111000110101111000011110111011111010100010011011100111010101101010101000000111001111101001001010011011001101111000000111111111111101001011011000001100110111100011100110010111001111001011100001100001100000001010100011010101111101111011010001100010000001101110100111111100111011010110101000110111101001011010100001101001010011110011111101010100010110111001000001110101001111000001100100110100100010110001111001100011111010000101011101001100001101110111111001111100010110101010110011011100101110100001001110011101101011011010100011100110101100000101001111101101000011110000110011011110100111110000001000010100011000110100000111100011101101000000111111101010000111101100100110001011111100011101010011001000001001000111011111101000110111011000110100110000010100001000100101010110011010011110100100000000001000101011000001010011011011110101110001001101110001100110101011001110010111000101001111101011011011000100100011100110100011000111111110100101000001101000000111101111101100000111011110111001101010011010011011010001100010010100101001000111111110001100100001001111010000101010101111011000111111011101110100110010100000001010000101011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1";
看到上面的这堆 0110 是不是有点懵逼？
没关系，我刚看到的时候也是一样的，用下面的代码把他转换成数组就好
List<int> students = new List<int>();
for (int i = 0; i < text.Length; i++)
{
students.Add(text[i] - 48);
}
你看，事情往往就是这么简单、有趣，一个 for 循环就完美解决了我们的麻烦。
方案一：暴力解题法
看到这个题目，我们常常第一时间就是想到循环，遍历找出这帮真男人（1）就好了不是？
来吧，下面就是一段暴力 for 循环
List<int> newStudents = new List<int>();
for (int i = 0; i < students.Count; i++)
{
if (students[i] == 1)
newStudents.Add(students[i]);
}
你看，这种以暴制暴的方式往往就是这么枯燥、无趣。
知识点来了！！！
由于我们声明了一个 newStudents 的泛型对象，这个对象在初始化的时候，默认大小是多少？正确答案是 4 。是的，你没看错，咱 .NETCore 就是这么扣，就是任性，就给你 4 个长度的列表。
那咱们就要问了，这 4 个是怎么装下这么多男人（1）的呢？
动态扩容
答案是：当我们调用 List.Add() 的时候，如果当前列表长度小于容量，即 Length < Capacity，那么数据就正常的装入列表中，如果即将装入的数据长度 + 现有的列表长度 > 容量，即 Length > Capacity，那么 List 底层就会进行自动扩容。
每次扩容多少合适呢？翻倍！
没错，就是这么简单有效，咱直接翻倍了，不装了，不算了，就是这么简单有效，咱就翻倍。
看 List 的底层扩容的实现
private void EnsureCapacity(int min) {
if (_items.Length < min) {
int newCapacity = _items.Length == 0? _defaultCapacity : _items.Length * 2;
// Allow the list to grow to maximum possible capacity (~2G elements) before encountering overflow.
// Note that this check works even when _items.Length overflowed thanks to the (uint) cast
if ((uint)newCapacity > Array.MaxArrayLength) newCapacity = Array.MaxArrayLength;
if (newCapacity < min) newCapacity = min;
Capacity = newCapacity;
}
}
重新分配内存扩容
看上面的代码，当调用 Capacity = newCapacity;进行扩容的时候，内存将进行重新分配，具体做法就是在属性 Capacity 内部，使用 newCapacity 声明一个新的数组，然后把源数组中的内容复制到新数组中。
public int Capacity {
get {
Contract.Ensures(Contract.Result<int>() >= 0);
return _items.Length;
}
set {
if (value < _size) {
ThrowHelper.ThrowArgumentOutOfRangeException(ExceptionArgument.value, ExceptionResource.ArgumentOutOfRange_SmallCapacity);
}
Contract.EndContractBlock();
if (value != _items.Length) {
if (value > 0) {
T[] newItems = new T[value];
if (_size > 0) {
Array.Copy(_items, 0, newItems, 0, _size);
}
_items = newItems;
}
else {
_items = _emptyArray;
}
}
}
}
所以，由此推断，我们一开始的暴力迭代方案将会在运行时发生多少次内存分配和 GC 呢，我们算一下，4/8/16/32/64/128/256/512/1024，算了，我算不过来了，反正就是很多很多！
暴力修正方案
从上面的知识点可以得知，我们应该一开始就应该声明一个合适长度的 List。
但是，我们又没有办法知道有多少男人（1）呀，那怎么办，这种时候就要逆向思维，声明一个足够大的数组，你源数据不是有 10000 条吗，那我就假定 10000 个都是男人（1），声明一个 10000 人的列表好了。
List<int> newStudents = new List<int>(students.Count);
for (int i = 0; i < students.Count; i++)
{
if (students[i] == 1)
newStudents.Add(students[i]);
}
完美！
方案二：算法实现
既然是算法题，如果只考察内存和GC，那未免显得我们学识粗浅，下面我们对数据进行排序，然后再截取数组实现解题。
冒泡排序
我们这里使用经典的冒泡排序法，当然你可以二分或者其它，随意发挥好了。
int total = 0;
for (int i = 0; i < students.Count; i++)
{
if (i == 1)
total++;
for (int j = 0; j < i; j++)
{
if (students[j] < students[i])
{
var stu = students[i];
students[i] = students[j];
students[j] = stu;
total++;
break;
}
}
}
上面的代码实现了对数据进行排序，排序的同时不要浪费机会，借机获取了男人（1）的数量，为下面的节省内存分配做准备，既然学了就要应用是吧。
截取数组
int[] newStudents = new int[total];
students.CopyTo(0, newStudents, 0, newStudents.Length);
截取数组的代码非常简单，首先利用排序得到的数量声明合适的列表，然后利用 List.CopyTo() ，方法即可得到男人（1）。
这里的 students.CopyTo() 底层实现顺便告诉大家，使用的是 Array.CopyTo()。是的，你没看错，咱
这是 List 的底层的实现
public void CopyTo(int index, T[] array, int arrayIndex, int count) {
if (_size - index < count) {
ThrowHelper.ThrowArgumentException(ExceptionResource.Argument_InvalidOffLen);
}
Contract.EndContractBlock();
// Delegate rest of error checking to Array.Copy.
Array.Copy(_items, index, array, arrayIndex, count);
}
本质上都是调用 Array 的方法。
方案三：修剪法
经过两轮磋商，相信这个时候面试官想必对你已经是刮目相看，别着急，下面还有一个方案，利用迭代列表，去修剪列表中的女人（0），剩下的自然都是男人（1）。
for (int i = 0; i < students.Count; i++)
{
if (students[i] == 0)
{
students.RemoveAt(i);
}
}
注意了，这是一段错误的代码！
注意了，这是一段错误的代码！
注意了，这是一段错误的代码！
上面的修剪代码虽然看似执行了修剪女人（0）的过程，实际上，我们要了解的是，在修剪的过程中， students.Count 会随着修剪的执行过程而同时会发生改变，最终结果就是，修剪长度被缩短，导致部分女人（1）没有得到修剪检查。
看 List.RemoveAt() 的内部实现。
public void RemoveAt(int index) {
if ((uint)index >= (uint)_size) {
ThrowHelper.ThrowArgumentOutOfRangeException();
}
Contract.EndContractBlock();
_size--;
if (index < _size) {
Array.Copy(_items, index + 1, _items, index, _size - index);
}
_items[_size] = default(T);
_version++;
}
List.Count 属性的返回值正式上文中的 _size；所以，你现在明白这个方案为什么是个错误了吧。
方案三的修正版本
那么我们就没有办法使用修剪方案了吗？
不是的，请年轻人先把40米的大刀收一收，我们来修正一下上面的修剪方案。
for (int i = 0; i < students.Count; i++)
{
if (students[i] == 0)
{
students.RemoveAt(i);
i--;
}
}
具体做法就是在执行修剪操作后，将当前下标 i 和 students.Count 进行同步，一个小小的改动，我们实现了修剪。
性能问题
通过上面的三种方案，我们学习了 List 的内部执行机制，了解了内存分配和GC的回收过程，那么这三种方案的性能怎么样呢？
是时候上图了。
总结
好了，今天的面试就到这里结束。
通过上面的性能对比，我们可以看到排序方案的耗时最高，耗时高达 166ms ，而其它两个方案的耗时都为 0，当然，从内存占用来看，排序方案肯定是要优秀一些的了，但是在这个求快的时代，内存这个可怜的家伙很多时候都会被我们安排在优化的第二梯队。
而从综合考虑，修剪方案较为可行（仅针对本题）。
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