二分法查找元素

基本原理

二分查找的思路很簡單，咱們設元素的開始和結尾的元素編號分別爲first和last。
除此以外，還須要設置另外的一個元素mid。
其中mid = (first+last)/2
二分法的實現思路是，每次查找都在以當前序列的中間值爲一個對比點，從而每次都會把查找範圍縮小到當前序列的一半的元素中。 下面是代碼實現：

//二分法查找
class Solutions2
{
public:
    searchBin(const vector<int>& nums,int target)
    {
        int first = 0;
        int last = nums.size();
        int mid = 0;
        while(first != last)
        {
            mid = (last+first)/2;
            if(nums[mid] == target)
                return mid;
            else
            {
                if(target > nums[mid])
                    first = mid + 1;
                if(target < nums[mid])
                    last = mid - 1;
            }
        }
        return -1;
    }
};

---

下面是測試代碼：

#include<iostream>
#include<vector>
using namespace std;

//二分法查找
class Solutions2
{
public:
    searchBin(const vector<int>& nums,int target)
    {
        int first = 0;
        int last = nums.size();
        int mid = 0;
        while(first != last)
        {
            mid = (last+first)/2;
            if(nums[mid] == target)
                return mid;
            else
            {
                if(target > nums[mid])
                    first = mid + 1;
                if(target < nums[mid])
                    last = mid - 1;
            }
        }
        return -1;
    }
};



//主函數
int main(void)
{

    vector<int> a;
    a.push_back(1);
    a.push_back(1);
    a.push_back(1);
    a.push_back(2);
    a.push_back(2);
    a.push_back(3);
    a.push_back(4);
    a.push_back(4);
    a.push_back(4);
    a.push_back(4);
    a.push_back(5);

    cout<<"after binary search "<<endl;
    Solutions2 ss;

    int return_binSort = ss.searchBin(a,3);
    cout <<"the result is :"<<return_binSort;
    cout<<endl;
    return 0;
}