演算法面試題：一個List
解題思路
這是群裡某位小夥伴去面試碰到的面試題，從題目本身來看，面試官應該是要考察面試者對泛型 List
首先，題目沒有限定數量，我們知道，只談毒性不談劑量的都是耍流氓，那我們就假定要處理的資料有 10000 條好了。
資料準備好了
static readonly string text = "1000100010101001101101011110111111100000100010001101010101001101111010100001010101100101011011101011111010101000111100110101001001000010100000010111010110011010111100000101110011011100100010110101100010111010100010110000001011110110000011110001101001000000111100110101101111111000011100010110010000000101111010111011100000101111000011101101110100111010101001010000001101000100111011111101010100011011000001001010100100001001011001100011001010111110110100101101101010010100010101011100110101110011011101001001001110101101011010101001000011001111010101110101010001111010110001011100010001000011000101000011010011011000100011111110111010001111111000100011110000100010010101000111110111111101111111000000001011010101000100100101010100000111101111010011011011100000001001000110011100110110001110100100110010011100000110111010010011100011001100010001011100010010001011100110101110000001000010001111101010110000111111110101110100111101110000111000110011110101001110010001010100110110010100011110110111010010011011100011010111011010100110011010001110011101000100010100011110001000101010100000010111011010101100001111011000001110110010011111011000010001101100001101100001010010000001001101110101110001100000100100100110101110110110101001111011011010111000100110011011111001010001100001011000110000100011000010010101011010100111001000110001011101010100010101111101100101001110100111000011110101111111001011100001001001100110011001010001111000110000101111010101010011101010111000000001001111001111101010101010101101100101010111001011100000010001011001000001110101000110111000011110101010110101101001001101001110000010000000010011111110101001111111000100100010001100101111011111000110100000000000110010011101010010011100000000111111000001011100000101100001100010101111010110000100011101101101111011100010011111100100100001001001011001000000010100110101110101101100001000011110001100010101011000100000000011101001101101100101010011000101100100001010100011100111110011100110110011100011100011110001101100000001111011100111110011111100001100001111101101111101110001111011010100011000101000111110110111011000011110010110010010100011100101111000101110010110100000011010000101110100001101001001110010110101101011110110001111101101000000011011010001111000010110000011110100101011011100100111001101001101110011101010001001001010010111100010111111111001101010000011000000000101111101011110111001001000010100000010100000110101001101100001111001011100000101010000010000111101100111011101001111000011110111011100011101110000101111000111011101010100000110011000010001101101110000111010111100001011110001101100001000110101010010001010000000001110111011110001110101000000101110010100110010110101011100000100010011100001001010100110100100101111010111100000101010100100101001010110000101110110010110100101101111000010100001010011111111010011101001001011001000101000010000000100111001011010110000010001110010011100010111111100101001010100111000100011001111100101100011100000101100110010101100110111100000001111111011000100010100110111011011001000010110100000001111011010001010101110010111101000010110001010110011001010100101010001110100110100111101010011011110011101011111100101011010101101010010000100110111000011011110110000010110110000001010101011110001110001101100110111011010001001010011110100011111110110110000000111100111100111001110111011011010010100101110001111001110001100110011111001001000110011101111111100111011001001010011101100010000111010111000111101010101100010100101000001000111000010001010110110100010100011010110111110111111101001101000010111000111111011001100001111110011111100000110100110000011010110100111001001100111001010011011011000111100001001111100100101101101010111101001101001000101000100101000010110010100001111001010010111101001100011100010111001110100111111101110011100000011110110100001100001011110110100011010010001100000011111110110001001101001010010010000001000111000110101111000011110111011111010100010011011100111010101101010101000000111001111101001001010011011001101111000000111111111111101001011011000001100110111100011100110010111001111001011100001100001100000001010100011010101111101111011010001100010000001101110100111111100111011010110101000110111101001011010100001101001010011110011111101010100010110111001000001110101001111000001100100110100100010110001111001100011111010000101011101001100001101110111111001111100010110101010110011011100101110100001001110011101101011011010100011100110101100000101001111101101000011110000110011011110100111110000001000010100011000110100000111100011101101000000111111101010000111101100100110001011111100011101010011001000001001000111011111101000110111011000110100110000010100001000100101010110011010011110100100000000001000101011000001010011011011110101110001001101110001100110101011001110010111000101001111101011011011000100100011100110100011000111111110100101000001101000000111101111101100000111011110111001101010011010011011010001100010010100101001000111111110001100100001001111010000101010101111011000111111011101110100110010100000001010000101011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1";
看到上面的這堆 0110 是不是有點懵逼？
沒關係，我剛看到的時候也是一樣的，用下面的程式碼把他轉換成陣列就好
List<int> students = new List<int>();
for (int i = 0; i < text.Length; i++)
{
students.Add(text[i] - 48);
}
你看，事情往往就是這麼簡單、有趣，一個 for 迴圈就完美解決了我們的麻煩。
方案一：暴力解題法
看到這個題目，我們常常第一時間就是想到迴圈，遍歷找出這幫真男人（1）就好了不是？
來吧，下面就是一段暴力 for 迴圈
List<int> newStudents = new List<int>();
for (int i = 0; i < students.Count; i++)
{
if (students[i] == 1)
newStudents.Add(students[i]);
}
你看，這種以暴制暴的方式往往就是這麼枯燥、無趣。
知識點來了！！！
由於我們宣告瞭一個 newStudents 的泛型物件，這個物件在初始化的時候，預設大小是多少？正確答案是 4 。是的，你沒看錯，我們 .NETCore 就是這麼扣，就是任性，就給你 4 個長度的列表。
那我們們就要問了，這 4 個是怎麼裝下這麼多男人（1）的呢？
動態擴容
答案是：當我們呼叫 List
每次擴容多少合適呢？翻倍！
沒錯，就是這麼簡單有效，我們直接翻倍了，不裝了，不算了，就是這麼簡單有效，我們就翻倍。
看 List
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;
}
}
重新分配記憶體擴容
看上面的程式碼，當呼叫 Capacity = newCapacity;進行擴容的時候，記憶體將進行重新分配，具體做法就是在屬性 Capacity 內部，使用 newCapacity 宣告一個新的陣列，然後把源陣列中的內容複製到新陣列中。
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，由此推斷，我們一開始的暴力迭代方案將會在執行時發生多少次記憶體分配和 GC 呢，我們算一下，4/8/16/32/64/128/256/512/1024，算了，我算不過來了，反正就是很多很多！
暴力修正方案
從上面的知識點可以得知，我們應該一開始就應該宣告一個合適長度的 List
但是，我們又沒有辦法知道有多少男人（1）呀，那怎麼辦，這種時候就要逆向思維，宣告一個足夠大的陣列，你源資料不是有 10000 條嗎，那我就假定 10000 個都是男人（1），宣告一個 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：演算法實現
既然是演算法題，如果只考察記憶體和GC，那未免顯得我們學識粗淺，下面我們對資料進行排序，然後再擷取陣列實現解題。
氣泡排序
我們這裡使用經典的氣泡排序法，當然你可以二分或者其它，隨意發揮好了。
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;
}
}
}
上面的程式碼實現了對資料進行排序，排序的同時不要浪費機會，藉機獲取了男人（1）的數量，為下面的節省記憶體分配做準備，既然學了就要應用是吧。
擷取陣列
int[] newStudents = new int[total];
students.CopyTo(0, newStudents, 0, newStudents.Length);
擷取陣列的程式碼非常簡單，首先利用排序得到的數量宣告合適的列表，然後利用 List
這裡的 students.CopyTo() 底層實現順便告訴大家，使用的是 Array.CopyTo()。是的，你沒看錯，我們
這是 List
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);
}
本質上都是呼叫 Array 的方法。
方案三：修剪法
經過兩輪磋商，相信這個時候面試官想必對你已經是刮目相看，彆著急，下面還有一個方案，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，這是一段錯誤的程式碼！
-
注意了，這是一段錯誤的程式碼！
-
注意了，這是一段錯誤的程式碼！
上面的修剪程式碼雖然看似執行了修剪女人（0）的過程，實際上，我們要了解的是，在修剪的過程中， students.Count 會隨著修剪的執行過程而同時會發生改變，最終結果就是，修剪長度被縮短，導致部分女人（1）沒有得到修剪檢查。
看 List
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
方案三的修正版本
那麼我們就沒有辦法使用修剪方案了嗎？
不是的，請年輕人先把40米的大刀收一收，我們來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--;
}
}
具體做法就是在執行修剪操作後，將當前下標 i 和 students.Count 進行同步，一個小小的改動，我們實現了修剪。
效能問題
通過上面的三種方案，我們學習了 List
是時候上圖了。
總結
好了，今天的面試就到這裡結束。
通過上面的效能對比，我們可以看到排序方案的耗時最高，耗時高達 166ms ，而其它兩個方案的耗時都為 0，當然，從記憶體佔用來看，排序方案肯定是要優秀一些的了，但是在這個求快的時代，記憶體這個可憐的傢伙很多時候都會被我們安排在優化的第二梯隊。
而從綜合考慮，修剪方案較為可行（僅針對本題）。