二分查找--C++实现

1. 简介

满足有序性，每次排除一半的可能性。

2. 实现

2.1 手写

int bin_search(vector<int> &arr,int v) {
	int hi = arr.size() - 1;
	int lo = 0;
	
	while ( lo <= hi)
	{
		int mid = (lo + hi) >> 1;
		if (arr[mid] < v)
			lo = mid + 1;
		else
			hi = mid - 1;
	}
	return hi;
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14

2.2 STL

1. lower_bound()
   找到第一不小于某个值的位置

• 使用

lower_bound(potions.begin(), potions.end(), tar);
1

• 实现

template<class ForwardIt, class T>
ForwardIt lower_bound(ForwardIt first, ForwardIt last, const T& value)
{
    ForwardIt it;
    typename std::iterator_traits<ForwardIt>::difference_type count, step;
    count = std::distance(first, last);
 
    while (count > 0)
    {
        it = first; 
        step = count / 2; 
        std::advance(it, step);
 
        if (*it < value)
        {
            first = ++it; 
            count -= step + 1; 
        }
        else
            count = step;
    }
 
    return first;
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24

2. upper_bound()
   找到第一个严格大于某个值的位置

• 使用

upper_bound(potions.begin(), potions.end(), tar);
1
2

• 实现

template<class ForwardIt, class T>
ForwardIt upper_bound(ForwardIt first, ForwardIt last, const T& value)
{
    ForwardIt it;
    typename std::iterator_traits<ForwardIt>::difference_type count, step;
    count = std::distance(first, last);
 
    while (count > 0)
    {
        it = first; 
        step = count / 2; 
        std::advance(it, step);
 
        if (!(value < *it))
        {
            first = ++it;
            count -= step + 1;
        } 
        else
            count = step;
    }
 
    return first;
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24

3. Ref

cppreference