二分搜尋法（搜尋原則的代表）

說明

如果搜尋的數列已經有排序，應該儘量利用它們已排序的特性，以減少搜尋比對的次數，這是搜尋的基本原則，二分搜尋法是這個基本原則的代表。

解法

在二分搜尋法中，從數列的中間開始搜尋，如果這個數小於我們所搜尋的數，由於數列已排序，則該數左邊的數一定都小於要搜尋的對象，所以無需浪費時間在左邊的數；如果搜尋的數大於所搜尋的對象，則右邊的數無需再搜尋，直接搜尋左邊的數。

所以在二分搜尋法中，將數列不斷的分為兩個部份，每次從分割的部份中取中間數比對，例如要搜尋92於以下的數列，首先中間數索引為(0+9)/2 = 4（索引由0開始）：

[3 24 57 57 67 68 83 90 92 95]

由於67小於92，所以轉搜尋右邊的數列：

3 24 57 57 67 [68 83 90 92 95]

由於90小於92，再搜尋右邊的數列，這次就找到所要的數了：

3 24 57 57 67 68 83 90 [92 95]

實作：C Java Python Scala Ruby

#include <stdio.h> 
#include <stdlib.h> 
#include <time.h> 
#define MAX 10 
#define SWAP(x,y) {int t; t = x; x = y; y = t;} 

void quickSort(int[], int, int); 
int binarySearch(int[], int); 

int main(void) { 
    srand(time(NULL)); 
    
    int number[MAX] = {0}; 

    int i;
    for(i = 0; i < MAX; i++) { 
        number[i] = rand() % 100; 
    } 

    quickSort(number, 0, MAX-1); 

    printf("數列："); 
    for(i = 0; i < MAX; i++) 
        printf("%d ", number[i]); 

    int find;
    printf("\n輸入尋找對象："); 
    scanf("%d", &find); 

    if((i = binarySearch(number, find)) >= 0) 
        printf("找到數字於索引 %d ", i); 
    else 
        printf("\n找不到指定數"); 
    
    printf("\n"); 

    return 0; 
} 

int binarySearch(int number[], int find) { 
    int low = 0; 
    int upper = MAX - 1; 
    while(low <= upper) { 
        int mid = (low+upper) / 2; 
        if(number[mid] < find) 
            low = mid+1; 
        else if(number[mid] > find) 
            upper = mid - 1; 
        else 
            return mid; 
    } 
    return -1; 
} 

void quickSort(int number[], int left, int right) { 
    if(left < right) { 
        int s = number[(left+right)/2]; 
        int i = left - 1; 
        int j = right + 1; 

        while(1) { 
            while(number[++i] < s) ;  // 向右找 
            while(number[--j] > s) ;  // 向左找 
            if(i >= j) 
                break; 
            SWAP(number[i], number[j]); 
        } 

        quickSort(number, left, i-1);   // 對左邊進行遞迴 
        quickSort(number, j+1, right);  // 對右邊進行遞迴 
    } 
}

Java

public class Search {
    public static int binary(int[] number, int des) {
        int low = 0; 
        int upper = number.length - 1; 

        while(low <= upper) { 
            int mid = (low+upper) / 2; 
            if(number[mid] < des) 
                low = mid+1; 
            else if(number[mid] > des) 
                upper = mid - 1; 
            else 
                return mid; 
        } 

        return -1; 
    }
    
    public static void main(String[] args) {
        int[] number = {1, 2, 3, 4, 6, 7, 8};
        int find = Search.binary(number, 2);
        System.out.println(find >= 0 ? "找到數值於索引" + find : "找不到數值"); 
    }
}

Python

def search(number, des):
    low = 0
    upper = len(number) - 1
    while low <= upper:
        mid = (low + upper) // 2
        if number[mid] < des:
            low = mid + 1
        elif number[mid] > des:
            upper = mid - 1
        else:
            return mid
    return -1

number = [1, 4, 2, 6, 7, 3, 9, 8]
number.sort()
find = search(number, 2)
print("找到數值於索引 " + str(find) if find >= 0 else "找不到數值")

Scala

object Search {
    def binary(number: Array[Int], des: Int) = {
        var low = 0
        var upper = number.length - 1
        var result = -1
        var isContinue = true
        while(isContinue && low <= upper) {
            val mid = (low + upper) / 2
            if(number(mid) < des) {
                low = mid + 1
            }
            else if(number(mid) > des) {
                upper = mid - 1
            }
            else {
                result = mid
                isContinue = false
            }
        }
        result
    }
}

val number = Array(1, 2, 3, 4, 6, 7, 8)
val find = Search.binary(number, 3)
println(if(find >= 0) "找到數值於索引 " + find else "找不到數值")

Ruby

# encoding: Big5
def search(number, des)
    low = 0
    upper = number.length - 1
    while low <= upper
        mid = (low + upper) / 2
        if number[mid] < des
            low = mid + 1
        elsif number[mid] > des
            upper = mid - 1
        else
            return mid
        end
    end
    -1
end

number = [1, 4, 2, 6, 7, 3, 9, 8]
number.sort!
find = search(number, 2)
print find >= 0 ? "找到數值於索引 " + find.to_s : "找不到數值", "\n"