定義:
二分法檢索的基本思想是設字典中的元素從小到大有序地存放在數組(array)中。首先將給定值key與字典中間位置上元素的關鍵碼(key)比較,如果相等,則檢索成功;否則,若key小,則在字典前半部分中繼續進行二分法檢索;若key大,則在字典后半部分中繼續進行二分法檢索。這樣,經過一次比較就縮小一半的檢索區間,如此進行下去,直到檢索成功或檢索失敗。偶數個取中間2個其中任何一個作為中間元素。二分法檢索是一種效率較高的檢索方法,要求字典在順序表中按關鍵碼排序。
例題:
大意為:
給定一個雙調數組(數組中數字先遞增后遞減),例如1,3,5,7,6,4,2,0,-2,-4,-6,-10。另外給定一個數字X。
設計一個算法,找出X是否在數組中。當此數組的數字總量n為很大的一個數值時,尋找X時,程序最多運行大約2lgN次。lg為底數為2的對數,lg4=2;lg8=3。
解答:
解題思路:首先,直接遍歷是肯定不行的,因為當n很大時,如果X是數組的最后一個數字,則程序要運行N次才能找到。例如:
int x; //如果x==RawArray[RawArray.Num()-1],則這個要遍歷所有數字,運行了N次 for (int i=0;i<RawArray.Num();++i) { if (RawArray[i] == x) return; }
然后,解題方法為:利用二分法檢索來檢索。先用二分法檢索找到遞增區域和遞減區域的共同數字(上述例子中的7);然后分別在遞增區域和遞減區域中用二分法檢索來尋找X。
具體實現:
.h: UCLASS() class ALGORITHM_API AAnalysisExerciseOne : public AActor { GENERATED_BODY() public: // Sets default values for this actor's properties AAnalysisExerciseOne(); //在數組中尋找X,如果找到,返回True;反之,返回false; bool FindIntX(TArray<int> TargetArray, int X); //二叉法找Int bool XInArray(TArray<int> TargetArray, int LeftIndex, int RightIndex, int X, bool IncreaseSide); protected: // Called when the game starts or when spawned virtual void BeginPlay() override; public: // Called every frame virtual void Tick(float DeltaTime) override; private: TArray<int> RawArray; }; .cpp: // Sets default values AAnalysisExerciseOne::AAnalysisExerciseOne() { // Set this actor to call Tick() every frame. You can turn this off to improve performance if you don't need it. PrimaryActorTick.bCanEverTick = true; } //在數組中尋找X,如果找到,返回True;反之,返回false; bool AAnalysisExerciseOne::FindIntX(TArray<int> TargetArray, int X) { //利用二叉法,尋找遞增數組和遞減數組的共用元素 //左右兩個點 int Left(0); int Right(TargetArray.Num()); //這個是要找的元素的index int TargetInt(0); //開始尋找 while (Left <= Right) { //找中間點,如果有小數點,舍棄,取整 int Mid = Left + (Right - Left) / 2; if (TargetArray[Mid - 1] < TargetArray[Mid] && TargetArray[Mid]>TargetArray[Mid + 1]) { //找到了,結束While TargetInt = Mid; break; } else if (TargetArray[Mid - 1] < TargetArray[Mid]) { Left = Mid + 1; } else if (TargetArray[Mid] > TargetArray[Mid + 1]) { Right = Mid - 1; } //在雙調數組中,TargetArray[Mid - 1] > TargetArray[Mid] && TargetArray[Mid] < TargetArray[Mid + 1]的情況不存在 else break; } //等於0說明沒找到,出問題了,返回 if (TargetInt == 0) return false; //X比數組中所有數字都大,X不在數組中 if (X > TargetArray[TargetInt]) return false; if (X == TargetArray[TargetInt]) return true; //先在增長數組中找 if (XInArray(TargetArray, 0, TargetInt - 1, X, true)) { UKismetSystemLibrary::PrintString(this, "Find in increase side!"); return true; } //如果在增長數組中找不到,則去減少數組中找 else { if (XInArray(TargetArray, TargetInt + 1, TargetArray.Num() - 1, X, false)) { UKismetSystemLibrary::PrintString(this, "Find in Decrease side!"); return true; } //都找不到,說明沒有 else { UKismetSystemLibrary::PrintString(this, "Don't Find it!"); return false; } } } bool AAnalysisExerciseOne::XInArray(TArray<int> TargetArray, int LeftIndex, int RightIndex, int X, bool IncreaseSide) { int Left(LeftIndex); int Right(RightIndex); //開始尋找 if (IncreaseSide) { //在遞增區域中尋找 while (Left <= Right) { //找中間點,如果有小數點,舍棄,取整 int Mid = Left + (Right - Left) / 2; if (X < TargetArray[Mid]) { Right = Mid - 1; } else if (X > TargetArray[Mid]) { Left = Mid + 1; } else return true; } return false; } //在遞減區域中尋找 else { while (Left <= Right) { //找中間點,如果有小數點,舍棄,取整 int Mid = Left + (Right - Left) / 2; if (X > TargetArray[Mid]) { Right = Mid - 1; } else if (X < TargetArray[Mid]) { Left = Mid + 1; } else return true; } return false; } } // Called when the game starts or when spawned void AAnalysisExerciseOne::BeginPlay() { Super::BeginPlay();
//測試 //給定一個初始數組 RawArray.Add(1); RawArray.Add(3); RawArray.Add(5); RawArray.Add(7); RawArray.Add(6); RawArray.Add(4); RawArray.Add(2); RawArray.Add(0); RawArray.Add(-2); RawArray.Add(-4); RawArray.Add(-6); RawArray.Add(-10); if (FindIntX(RawArray, -10)) { UKismetSystemLibrary::PrintString(this, "Find it!"); } else { UKismetSystemLibrary::PrintString(this, "Don't Find it!"); } } // Called every frame void AAnalysisExerciseOne::Tick(float DeltaTime) { Super::Tick(DeltaTime); }