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]
#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
}
}
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");
}
}
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")
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")
# 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"
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