一.起緣
故事緣于一位朋友的一道題：
朋友四人玩LOL游戲。第一局，分別選擇位置：中單，上單，ADC，輔助；第二局新加入的伙伴要選上單，四人可選位置變?yōu)椋褐袉危蛞埃珹DC，輔助；要求，第二局四人每人不得選擇和第一局相同的位置，請問兩局綜合考慮有多少種位置選擇方式？
對于像我這邊不懂游戲的人來講，看不懂。于是有了這個(gè)版本：
有4個(gè)人，4只椅子，第一局每人坐一只椅子，第二局去掉第2只椅子，增加第5只椅子，每人坐一只椅子，而且每個(gè)人不能與第一局坐相同的椅子。問兩局綜合考慮，共有多少種可能的情況？
我一開始的想法是這樣的，4個(gè)人就叫ABCD：第1局可能數(shù)是4*3*2*1=24，如果A第1局選了第2張椅，則A有4種可能，否則A有3種可能。對B來講，如果A選了B第一局的椅，則B有3種可能，否則B有2種可能（排隊(duì)自己第一局和A第二局已選）……想到這里我就暈了，情況越分越多。
二.原始的for嵌套
本來是一道數(shù)學(xué)題，應(yīng)該由知識(shí)算出來有多少種，但我突然有個(gè)想法，不如用計(jì)算機(jī)窮舉出出來。一來可以為各種猜測提供一個(gè)正確的答案，二來或許可以從答案反推出(數(shù)學(xué)上的)計(jì)算方法。然后就寫了第1版：
static Seat data = new Seat();public static void Run(){for (int a = 0; a < 4; a++){if (data.IsSelected(0, a)) //第1局編號(hào)0。如果已經(jīng)被人坐了。continue;data.Selected(0, a, "A"); //第1局編號(hào)0。A坐a椅。for (int b = 0; b < 4; b++){if (data.IsSelected(0, b))continue;data.Selected(0, b, "B");for (int c = 0; c < 4; c++){if (data.IsSelected(0, c))continue;data.Selected(0, c, "C");for (int d = 0; d < 4; d++){if (data.IsSelected(0, d))continue;data.Selected(0, d, "D");for (int a2 = 0; a2 < 5; a2++){if (a2 == 1)continue;if (data.IsSelected(1, a2)) //第2局編號(hào)1continue;if (data.IsSelected(0, a2, "A")) //如果第1局A坐了a2椅continue;data.Selected(1, a2, "A");for (int b2 = 0; b2 < 5; b2++){if (b2 == 1)continue;if (data.IsSelected(1, b2))continue;if (data.IsSelected(0, b2, "B"))continue;data.Selected(1, b2, "B");for (int c2 = 0; c2 < 5; c2++){if (c2 == 1)continue;if (data.IsSelected(1, c2))continue;if (data.IsSelected(0, c2, "C"))continue;data.Selected(1, c2, "C");for (int d2 = 0; d2 < 5; d2++){if (d2 == 1)continue;if (data.IsSelected(1, d2))continue;if (data.IsSelected(0, d2, "D"))continue;data.Selected(1, d2, "D");data.Count++; //可能的情況數(shù)加1Console.WriteLine("{0,5} {1}", data.Count, data.Current);data.UnSelected(1, d2);}data.UnSelected(1, c2);}data.UnSelected(1, b2);}data.UnSelected(1, a2);}data.UnSelected(0, d);}data.UnSelected(0, c);}data.UnSelected(0, b);}data.UnSelected(0, a); //A起身（釋放坐椅)}} 部分運(yùn)行結(jié)果：
說明：
1.ABCD是人名
2.“.”代表沒有人
3.位置是是座位
4.-左邊是第1局，右邊是第2局
5.數(shù)字是序號(hào)
1 A B C D .-B . A C D
2 A B C D .-C . A B D
3 A B C D .-D . A B C
4 A B C D .-D . A C B
5 A B C D .-B . D A C
6 A B C D .-C . B A D
7 A B C D .-D . B A C
8 A B C D .-C . D A B
9 A B C D .-B . D C A
10 A B C D .-D . B C A
11 A B C D .-C . D B A
12 A B D C .-B . A D C
...
262 D C B A .-B . C D A
263 D C B A .-B . D C A
264 D C B A .-C . D B A
算出來是264種。從答案上來看是每11種是一組，一組中第1局的坐法是相同的，也就是說對于第一局的每一種情況，第2局都是有11種不同的可能。而第一局的可能性是24，所以答案是24*11=264。而第2局為什么是11種可能，后面再說。
三.想要鏈?zhǔn)綄懛?/span>
主題來了，像剛才的第1版的寫法太死板太麻煩了。
如果能像這樣寫代碼就爽了：
obj.Try("A").Try("B").Try("C").Try("D").Try2("A").Try2("B").Try2("C").Try2("D").Write(); 而這樣的代碼通常的邏輯是執(zhí)行Try("A")方法，然后執(zhí)行Try("A")它return的對象的Try("B")方法……，即是Try("B")方法只被執(zhí)行1次，而我希望的是Try("B")方法被Try("A")內(nèi)部循環(huán)調(diào)用n次，Try("C")方法又被Try("B")方法調(diào)用m次。想想第1版的for套for不難明白為什么要追求這樣的效果。如果Try("A")執(zhí)行完了，再去執(zhí)行Try("B")，那么Try("B")肯定不會(huì)被調(diào)用多次，所以得延遲Try("A")的執(zhí)行，同理也延遲所有Try和Try2的執(zhí)行。由于lambda表達(dá)天生有延遲計(jì)算的特性，于是很快寫出了第2版：
public static void Run2(){Try("A",() => Try("B",() => Try("C",() => Try("D",() => Try2("A",() => Try2("B",() => Try2("C",() => Try2("D",null))))))));}public static void Try(string name, Action action) //第1局{for (int i = 0; i < 4; i++){if (data.IsSelected(0, i))continue;data.Selected(0, i, name);if (action == null){Console.WriteLine(data.Current);}else{action();}data.UnSelected(0, i);}}public static void Try2(string name, Action action) //第2局{for (int i = 0; i < 5; i++){if (i == 1)continue;if (data.IsSelected(1, i))continue;if (data.IsSelected(0, i, name))continue;data.Selected(1, i, name);if (action == null){data.Count++;Console.WriteLine("{0,5} {1}", data.Count, data.Current);}else{action();}data.UnSelected(1, i);}} 結(jié)構(gòu)更合理，邏輯更清晰，但是一堆lambda嵌套，太丑了，也不是我要的效果，我要的是類似這樣的：
obj.Try("A").Try("B").Try("C").Try("D").Try2("A").Try2("B").Try2("C").Try2("D").Write();
四.繼續(xù)向目標(biāo)逼近。
由于要延遲，所以必須先把要被調(diào)用的方法的引用“告訴”上一級(jí)，當(dāng)上一級(jí)執(zhí)行for的時(shí)候，就能調(diào)用下一級(jí)的方法。于是我想到了一個(gè)“回調(diào)鏈”
所以，執(zhí)行鏈?zhǔn)椒椒ㄊ窃跇?gòu)造回調(diào)鏈，最后的方法再通過調(diào)用鏈頭(Head)的某個(gè)方法啟動(dòng)真正要執(zhí)行的整個(gè)邏輯。
延遲計(jì)算是從Linq借鑒和學(xué)習(xí)來的，構(gòu)造Linq的過程并沒有執(zhí)行，等到了執(zhí)行ToList, First等方法時(shí)才真正去執(zhí)行。
我想構(gòu)造回調(diào)鏈每一步都是一個(gè)固定的方法，這里隨便起用了T這個(gè)極短名稱，而每一步后期計(jì)算時(shí)要執(zhí)行的方法可靈活指定。于是有了第3版：
static Seat data = new Seat(); //借用Seat保存數(shù)據(jù)public Seat2(string name, Seat2 parent, Action<Seat2> method){this.Name = name;this.Parent = parent;if (parent != null)parent.Child = this;this.Method = method;}public static void Run(){new Seat2("A", null, me => me.Try()).T("B", me => me.Try()).T("C", me => me.Try()).T("D", me => me.Try()).T("A", me => me.Try2()).T("B", me => me.Try2()).T("C", me => me.Try2()).T("D", me => me.Try2()).P().Start();}public Seat2 T(string name, Action<Seat2> method){return new Seat2(name, this, method);}public void Try(){for (int i = 0; i < 4; i++){if (data.IsSelected(0, i))continue;data.Selected(0, i, this.Name);if (this.Child != null){this.Child.Method(this.Child);}data.UnSelected(0, i);}}public void Try2(){for (int i = 0; i < 5; i++){if (i == 1)continue;if (data.IsSelected(1, i))continue;if (data.IsSelected(0, i, this.Name))continue;data.Selected(1, i, this.Name);if (this.Child != null){this.Child.Method(this.Child);}data.UnSelected(1, i);}} 五.解耦
這種調(diào)用方式，是滿意了。但是運(yùn)算框架與具體的算法耦合在一起，如果能把運(yùn)算框架提取出來，以后寫具體的算法也方便許多。于是經(jīng)過苦逼的提取，測試，踩坑，最終出現(xiàn)了第4版：
//運(yùn)算框架class ComputeLink<T> where T : ISeat{ComputeLink<T> Parent { get; set; } //父節(jié)點(diǎn)，即上一級(jí)節(jié)點(diǎn)ComputeLink<T> Child { get; set; } //子節(jié)點(diǎn)，即下一級(jí)節(jié)點(diǎn)T Obj { get; set; } //當(dāng)前節(jié)點(diǎn)對應(yīng)的算法對象，可以看作業(yè)務(wù)對象public ComputeLink(T obj, ComputeLink<T> parent, Action<T> method){if (obj == null)throw new ArgumentNullException("obj");this.Obj = obj;this.Obj.Method = x => method((T)x);if (parent != null){this.Parent = parent;parent.Child = this;parent.Obj.Child = this.Obj;}}public static ComputeLink<T> New(T obj, Action<T> method){return new ComputeLink<T>(obj, null, method);}public ComputeLink<T> Do(T obj, Action<T> method){return new ComputeLink<T>(obj, this, method);}public ComputeLink<T> Head //鏈表的頭{get{if (null != this.Parent)return this.Parent.Head;return this;}}public void Action() //啟動(dòng)(延遲的)整個(gè)計(jì)算{var head = this.Head;head.Obj.Method(head.Obj);}}interface ISeat{ISeat Child { get; set; }Action<ISeat> Method { get; set; }} p.s.為什么第4版是ISeat3而不是ISeat4呢，因?yàn)槲冶静话训?版當(dāng)作1個(gè)版本，因?yàn)樘剂耍霈F(xiàn)第2版后，我就把第1版給刪除了。為了寫這篇文章才重新去寫第1版。于是原本我當(dāng)作第3版的ISeat3自然地排到了第4版。
具體的"算法"就很簡單了：
class Seat3 : ISeat3{static Seat data = new Seat();string Name { get; set; }public Seat3(string name){this.Name = name;}/// <summary>/// 解耦的版本/// </summary>public static void Run(){var sql = ComputeLink<Seat3>.New(new Seat3("A"), m => m.Try()).Do(new Seat3("B"), m => m.Try()).Do(new Seat3("C"), m => m.Try()).Do(new Seat3("D"), m => m.Try()).Do(new Seat3("A"), m => m.Try2()).Do(new Seat3("B"), m => m.Try2()).Do(new Seat3("C"), m => m.Try2()).Do(new Seat3("D"), m => m.Try2()).Do(new Seat3(""), m => m.Print());sql.Action();}public Action<ISeat3> Method { get; set; }public ISeat3 Child { get; set; }public void Try(){for (int i = 0; i < 4; i++){if (data.IsSelected(0, i))continue;data.Selected(0, i, this.Name);if (this.Child != null){this.Child.Method(this.Child);}data.UnSelected(0, i);}}public void Try2(){for (int i = 0; i < 5; i++){if (i == 1)continue;if (data.IsSelected(1, i))continue;if (data.IsSelected(0, i, this.Name))continue;data.Selected(1, i, this.Name);if (this.Child != null){this.Child.Method(this.Child);}data.UnSelected(1, i);}}public void Print(){data.Count++;Console.WriteLine("{0,5} {1}", data.Count, data.Current);}} Seat3寫起來簡單，(Run方法內(nèi)部)看起來舒服。通過鏈?zhǔn)綄懛ㄟ_(dá)到嵌套循環(huán)的效果。對，這就是我要的！
它很像linq，所以我直接給變量命名為sql。
•對于Try和Try2來講，要調(diào)用的方法最好從參數(shù)傳來，但是這樣就會(huì)增加Run方法中New和Do的參數(shù)復(fù)雜性，破壞了美感，所以經(jīng)過權(quán)衡，Child和Method通過屬性傳入。這個(gè)我也不確定這樣做好不好，請各位大俠指點(diǎn)。
•還有一個(gè)細(xì)節(jié)，就是ComputeLink構(gòu)造方法中的（行號(hào)12的）代碼 this.Obj.Method = x => method((T)x); 。我原來是這樣寫的 this.Obj.Method = method; 編譯不通過，原因是不能把 Action<ISeat3> 轉(zhuǎn)化為 Action<T> ，雖然T一定實(shí)現(xiàn)了ISeat3，強(qiáng)制轉(zhuǎn)化也不行，想起以前看過的一篇文章里面提到希望C#以后的版本能擁有的一特性叫“協(xié)變”，很可能指的就是這個(gè)。既然這個(gè) Action<ISeat3> 不能轉(zhuǎn)化為 Action<T> 但是ISeat3是可以強(qiáng)制轉(zhuǎn)化為T的，所以我包了一層薄薄的殼，成了 this.Obj.Method = x => method((T)x); ，如果有更好的辦法請告訴我。
六.第2局為什么是11種可能
回過頭來解決為什么對于一個(gè)確定的第1局，第2局有11種可能。
不妨假設(shè)第1局的選擇是A選1號(hào)椅，B選2號(hào)椅，C選3號(hào)椅，D選4號(hào)椅。
第2局分為兩大類情況：
如果B選了第5號(hào)椅
則只有2種可能：
A B C D .-D . A C B
A B C D .-C . D A B
如果B選了不是第5號(hào)椅，
則ACD都有可能選第5號(hào)椅，有3種可能。B有3種選的可能(1,3,4號(hào)椅)，B一旦確定，A和C也只有一種可能
所以11 = 2 + 3 * 3
七.結(jié)論
由一道數(shù)學(xué)題牽引出多層循環(huán)嵌套，最終通過封裝達(dá)到了我要的鏈?zhǔn)秸{(diào)用的效果，我是很滿意的。這也是我第一次設(shè)計(jì)延遲計(jì)算，感覺強(qiáng)烈。如果新的場景需要用到延遲計(jì)算我想有了這次經(jīng)驗(yàn)寫起來會(huì)順手許多。如果是需要多層for的算法題都可以比較方便的實(shí)現(xiàn)了。
你都看到這里了，為我點(diǎn)個(gè)贊吧，能說一下看法就更好了。
完整代碼：
using System;using System.Linq;using System.Diagnostics;namespace ConsoleApplication{class Seat{static Seat data = new Seat();public static void Run(){//Seat.Run();//return;for (int a = ; a < ; a++){if (data.IsSelected(, a)) //第局編號(hào)。如果已經(jīng)被人坐了。continue;data.Selected(, a, "A"); //第局編號(hào)。A坐a椅。for (int b = ; b < ; b++){if (data.IsSelected(, b))continue;data.Selected(, b, "B");for (int c = ; c < ; c++){if (data.IsSelected(, c))continue;data.Selected(, c, "C");for (int d = ; d < ; d++){if (data.IsSelected(, d))continue;data.Selected(, d, "D");for (int a = ; a < ; a++){if (a == )continue;if (data.IsSelected(, a)) //第局編號(hào)continue;if (data.IsSelected(, a, "A")) //如果第局A坐了a椅continue;data.Selected(, a, "A");for (int b = ; b < ; b++){if (b == )continue;if (data.IsSelected(, b))continue;if (data.IsSelected(, b, "B"))continue;data.Selected(, b, "B");for (int c = ; c < ; c++){if (c == )continue;if (data.IsSelected(, c))continue;if (data.IsSelected(, c, "C"))continue;data.Selected(, c, "C");for (int d = ; d < ; d++){if (d == )continue;if (data.IsSelected(, d))continue;if (data.IsSelected(, d, "D"))continue;data.Selected(, d, "D");data.Count++; //可能的情況數(shù)加Console.WriteLine("{,} {}", data.Count, data.Current);data.UnSelected(, d);}data.UnSelected(, c);}data.UnSelected(, b);}data.UnSelected(, a);}data.UnSelected(, d);}data.UnSelected(, c);}data.UnSelected(, b);}data.UnSelected(, a); //A起身（釋放坐椅)}}public static void Run(){Try("A",() => Try("B",() => Try("C",() => Try("D",() => Try("A",() => Try("B",() => Try("C",() => Try("D",null))))))));}public static void Try(string name, Action action){for (int i = ; i < ; i++){if (data.IsSelected(, i))continue;data.Selected(, i, name);if (action == null){Console.WriteLine(data.Current);}else{action();}data.UnSelected(, i);}}public static void Try(string name, Action action){for (int i = ; i < ; i++){if (i == )continue;if (data.IsSelected(, i))continue;if (data.IsSelected(, i, name))continue;data.Selected(, i, name);if (action == null){data.Count++;Console.WriteLine("{,} {}", data.Count, data.Current);}else{action();}data.UnSelected(, i);}}public Seat(){seats[, ] = ".";seats[, ] = ".";}private string[,] seats = new string[, ];public void UnSelected(int game, int i){Debug.Assert(game == && i != || game == && i != );Debug.Assert(seats[game, i] != null);seats[game, i] = null;}public void Selected(int game, int i, string name){Debug.Assert(game == && i != || game == && i != );Debug.Assert(seats[game, i] == null);seats[game, i] = name;}public bool IsSelected(int game, int a){return seats[game, a] != null && seats[game, a] != ".";}public bool IsSelected(int game, int a, string name){return seats[game, a] == name;}public string Current{get{return string.Format("{} {} {} {} {}-{} {} {} {} {}",seats[, ], seats[, ], seats[, ], seats[, ], seats[, ],seats[, ], seats[, ], seats[, ], seats[, ], seats[, ]);}}public int Count { get; set; }}class Seat{static Seat data = new Seat(); //借用Seat保存法的數(shù)據(jù)Seat Parent { get; set; }Seat Child { get; set; }string Name { get; set; }Action<Seat> Method { get; set; }public Seat(string name, Seat parent, Action<Seat> method){this.Name = name;this.Parent = parent;if (parent != null)parent.Child = this;this.Method = method;}/// <summary>/// 耦合的版本/// </summary>public static void Run(){new Seat("A", null, me => me.Try()).T("B", me => me.Try()).T("C", me => me.Try()).T("D", me => me.Try()).T("A", me => me.Try()).T("B", me => me.Try()).T("C", me => me.Try()).T("D", me => me.Try()).P().Start();}public Seat T(string name, Action<Seat> method){return new Seat(name, this, method);}public Seat P(){return new Seat("Print", this, me => me.Print());}public void Start(){var head = this.Head;head.Method(head);}public Seat Head{get{if (null != this.Parent)return this.Parent.Head;return this;}}public void Try(){for (int i = ; i < ; i++){if (data.IsSelected(, i))continue;data.Selected(, i, this.Name);if (this.Child != null){this.Child.Method(this.Child);}data.UnSelected(, i);}}public void Try(){for (int i = ; i < ; i++){if (i == )continue;if (data.IsSelected(, i))continue;if (data.IsSelected(, i, this.Name))continue;data.Selected(, i, this.Name);if (this.Child != null){this.Child.Method(this.Child);}data.UnSelected(, i);}}public void Print(){data.Count++;Console.WriteLine("{,} {}", data.Count, data.Current);}public override string ToString(){return this.Name.ToString();}}class ComputeLink<T> where T : ISeat{ComputeLink<T> Parent { get; set; } //父節(jié)點(diǎn)，即上一級(jí)節(jié)點(diǎn)ComputeLink<T> Child { get; set; } //子節(jié)點(diǎn)，即下一級(jí)節(jié)點(diǎn)T Obj { get; set; } //當(dāng)前節(jié)點(diǎn)對應(yīng)的算法對象，可以看作業(yè)務(wù)對象public ComputeLink(T obj, ComputeLink<T> parent, Action<T> method){if (obj == null)throw new ArgumentNullException("obj");this.Obj = obj;this.Obj.Method = x => method((T)x);if (parent != null){this.Parent = parent;parent.Child = this;parent.Obj.Child = this.Obj;}}public static ComputeLink<T> New(T obj, Action<T> method){return new ComputeLink<T>(obj, null, method);}public ComputeLink<T> Do(T obj, Action<T> method){return new ComputeLink<T>(obj, this, method);}public ComputeLink<T> Head //鏈表的頭{get{if (null != this.Parent)return this.Parent.Head;return this;}}public void Action() //啟動(dòng)(延遲的)整個(gè)計(jì)算{var head = this.Head;head.Obj.Method(head.Obj);}}interface ISeat{ISeat Child { get; set; }Action<ISeat> Method { get; set; }}class Seat : ISeat{static Seat data = new Seat();string Name { get; set; }public Seat(string name){this.Name = name;}/// <summary>/// 解耦的版本/// </summary>public static void Run(){var sql = ComputeLink<Seat>.New(new Seat("A"), m => m.Try()).Do(new Seat("B"), m => m.Try()).Do(new Seat("C"), m => m.Try()).Do(new Seat("D"), m => m.Try()).Do(new Seat("A"), m => m.Try()).Do(new Seat("B"), m => m.Try()).Do(new Seat("C"), m => m.Try()).Do(new Seat("D"), m => m.Try()).Do(new Seat(""), m => m.Print());sql.Action();}public Action<ISeat> Method { get; set; }public ISeat Child { get; set; }public void Try(){for (int i = ; i < ; i++){if (data.IsSelected(, i))continue;data.Selected(, i, this.Name);if (this.Child != null){this.Child.Method(this.Child);}data.UnSelected(, i);}}public void Try(){for (int i = ; i < ; i++){if (i == )continue;if (data.IsSelected(, i))continue;if (data.IsSelected(, i, this.Name))continue;data.Selected(, i, this.Name);if (this.Child != null){this.Child.Method(this.Child);}data.UnSelected(, i);}}public void Print(){data.Count++;Console.WriteLine("{,} {}", data.Count, data.Current);}public override string ToString(){return this.Name.ToString();}}} 以上內(nèi)容是小編給大家介紹的C#用鏈?zhǔn)椒椒ū磉_(dá)循環(huán)嵌套的相關(guān)知識(shí)，希望對大家有所幫助！
