Question

我正在研究迭代器，并且已经停留了3天，弄清楚我们为什么要使用它们：

auto mid = text.begin() + (end - beg) / 2;

代码：

int main()

{
    vector<int> text{ 10,9,8,7,6,5,4,3,2,1 };
    int sought = 3;
    // text must be sorted
    // beg and end will denote the range we're searching
    auto beg = text.begin(), end = text.end();
    auto mid = text.begin() + (end - beg) / 2; // original midpoint
                                               // while there are still elements to look at and we haven't yet found sought
    while (mid != end && *mid != sought) {
        if (sought < *mid) // is the element we want in the first half?
            end = mid; // if so, adjust the range to ignore the second half
        else // the element we want is in the second half
            beg = mid + 1; // start looking with the element just after mid
        mid = beg + (end - beg) / 2;// new midpoint
    }

    system("pause");
}

为什么

auto mid = text.begin() + (end - beg) / 2;

而不是：

auto mid = text.begin() + text.size() / 2;

请帮忙。

Answer 1

这样做是为了避免在添加两个非常大的整数时可能发生的溢出，其中加法结果可能变得大于最大整数限制并产生奇怪的结果。

Extra, Extra - Read All About It: Nearly All Binary Searches and Mergesorts are Broken

来自博客：

So what's the best way to fix the bug? Here's one way:
 6:             int mid = low + ((high - low) / 2);

Probably faster, and arguably as clear is:
 6:             int mid = (low + high) >>> 1;

In C and C++ (where you don't have the >>> operator), you can do this:
 6:             mid = ((unsigned int)low + (unsigned int)high)) >> 1;

Answer 2

二进制搜索传统上是这样编写的。这种写作形式有助于编码人员理解二进制搜索，因为在标准二进制搜索中只使用了开始，结束，中间。

你可以}而不是size() 在循环之前，但你必须使用end-star循环，因为end-start会改变。您应该避免使用end-start来保持一致性。

使用迭代器进行二进制搜索，为什么我们使用“（end-begin）/ 2”？

2 个答案: