Note: The prerequisite for searching must be an ordered array or container
thought:
Define llow as the leftmost element position of the sequence table, and high as the right element position of the sequence table. Define mid = (low+high) / 2, that is, the middle position of the sequence table, and then compare the searched value with the value at the position of mid. Because the list is ordered, if the searched value is smaller than mid, only Search in the first half of the table, otherwise you only need to search in the second half of the table (if the two values are found to be equal in the first comparison, the current value will be returned directly), and so on, until the value you are looking for is found or Make sure that the value you are looking for is not in the list (that is, the search fails).
When there are no duplicate elements in the ordered array
#include<iostream>
using namespace std;
//二分查找算法---返回查找到的元素的下标
//数组 数组长度 查找的值
int test(int arr[],int len,int val)
{
//判断传入的数组值是否合法
if (len < 1)
{
return -1;
}
//最小值和最大值
int low = 0;
int high = len - 1;
//
while (low <= high)
{
int mid = (low + high) / 2;
//判断查找元素值比中间元素值大还是小
if (val > arr[mid])
{
low = mid + 1; //那么要去比mid大的左边区间进行折半查找，需要把最小值更新到mid+1
}
if (val < arr[mid])
{
high = mid - 1;//那么要去比mid小的右边区间进行折半查找，需要把最大值更新到mid-1
}
if (val == arr[mid])
{
return mid; //返回的就是查找到的元素下标
}
}
}
int main()
{
int arr[8] = {
-1,0,1,2,3,4,10,12};
int num=test(arr,8,4);
cout << "4元素在数组中下标位置为：" << num << endl;
system("pause");
return 0;
}
If there are multiple consecutively repeated values in an array, what should we do if we want to find the smallest subscript among the consecutively repeated values? We only need to slightly modify the else statement
#include<iostream>
using namespace std;
//二分查找算法---返回查找到的元素的下标
//数组 数组长度 查找的值
int test(int arr[],int len,int val)
{
//判断传入的数组值是否合法
if (len < 1)
{
return -1;
}
//最小值和最大值
int low = 0;
int high = len - 1;
//
while (low <= high)
{
int mid = (low + high) / 2;
//判断查找元素值比中间元素值大还是小
if (val > arr[mid])
{
low = mid + 1; //那么要去比mid大的左边区间进行折半查找，需要把最小值更新到mid+1
}
if (val < arr[mid])
{
high = mid - 1;//那么要去比mid小的右边区间进行折半查找，需要把最大值更新到mid-1
}
if (val == arr[mid])
{
//mid值要大于最小值的下标，判断mid前面是否有与val值相等的元素
while (mid > low && arr[mid - 1] == val)//在不越界的前提下后一个值与前一个值相等
{
--mid;//下标减一
}
return mid; //返回的就是查找到的元素下标
}
}
}
int main()
{
int arr[9] = {
-1,1,1,2,3,4,4,10,12};
int num=test(arr,8,4);
cout << "4元素在数组中下标位置为：" << num << endl;
system("pause");
return 0;
}
Recursive method implementation
#include<iostream>
using namespace std;
//二分查找算法---返回查找到的元素的下标
//数组 数组长度 查找的值
int test(int arr[],int len,int val,int low,int high)
{
//判断传入的数组值是否合法
if (len < 1)
{
return -1;
}
//
int mid = -1;
//
if (low <= high)
{
mid = (low + high) / 2;
//判断查找元素值比中间元素值大还是小
if (val > arr[mid])//要去mid右边，即比左边比mid大的区间
{
//这里如果在大区间里面找到了val的位置，就会返回下标，用mid来接收
//没有找到返回-1
mid=test(arr, len, val, mid + 1, high);
}
if (val < arr[mid])//要去mid左边，即比左边比mid小的区间
{
//这里如果在小区间里面找到了val的位置，就会返回下标，用mid来接收
mid=test(arr, len, val, low, mid - 1);
}
if (val == arr[mid])
{
while (mid > low && arr[mid - 1] == val)//在不越界的前提下后一个值与前一个值相等
{
--mid;//下标减一
}
return mid; //返回的就是查找到的元素下标
}
}
//元素不在当前区间数组中
return mid;
}
int main()
{
int arr[9] = {
-1,1,1,2,3,4,4,10,12};
int num=test(arr,9,4,0,8);
cout << "10元素在数组中下标位置为：" << num << endl;
system("pause");
return 0;
}
