blocks|key|1498662|text|您知道，这不是一个站点，您需要有人为您编写代码，因此，以下面的搜索为例，让您的方式：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1498663|//+binary_search+example
#include+<iostream>+++++//+std::cout
#include+<algorithm>++++//+std::binary_search,+std::sort
#include+<vector>+++++++//+std::vector

bool+myfunction+(int+i,int+j)+{+return+(i<j);+}

int+main+()+{
++int+myints[]+=+{1,2,3,4,5,4,3,2,1};
++std::vector<int>+v(myints,myints%2B9);+++++++++++++++++++++++++//+1+2+3+4+5+4+3+2+1

++//+using+default+comparison:
++std::sort+(v.begin(),+v.end());

++std::cout+<<+"looking+for+a+3...+";
++if+(std::binary_search+(v.begin(),+v.end(),+3))
++++std::cout+<<+"found!\n";+else+std::cout+<<+"not+found.\n";

++//+using+myfunction+as+comp:
++std::sort+(v.begin(),+v.end(),+myfunction);

++std::cout+<<+"looking+for+a+6...+";
++if+(std::binary_search+(v.begin(),+v.end(),+6,+myfunction))
++++std::cout+<<+"found!\n";+else+std::cout+<<+"not+found.\n";

++return+0;
}|code-block|syntax|javascript|1498664|正如Piotr所说，这不会给你索引，而是一个是/否的答案。然而，应该是最简单的方法，因此最快。|1498665|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cplusplus.com/reference/algorithm/binary_search/^0|V|2|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

You know SO is not a site that you request someone to write you the code, so take this example of <a href="http://www.cplusplus.com/reference/algorithm/binary_search/" rel="nofollow">std::binary_search</a> and make your way:

<pre><code>// binary_search example
#include &lt;iostream&gt; // std::cout
#include &lt;algorithm&gt; // std::binary_search, std::sort
#include &lt;vector&gt; // std::vector

bool myfunction (int i,int j) { return (i&lt;j); }

int main () {
 int myints[] = {1,2,3,4,5,4,3,2,1};
 std::vector&lt;int&gt; v(myints,myints+9); // 1 2 3 4 5 4 3 2 1

 // using default comparison:
 std::sort (v.begin(), v.end());

 std::cout &lt;&lt; "looking for a 3... ";
 if (std::binary_search (v.begin(), v.end(), 3))
 std::cout &lt;&lt; "found!\n"; else std::cout &lt;&lt; "not found.\n";

 // using myfunction as comp:
 std::sort (v.begin(), v.end(), myfunction);

 std::cout &lt;&lt; "looking for a 6... ";
 if (std::binary_search (v.begin(), v.end(), 6, myfunction))
 std::cout &lt;&lt; "found!\n"; else std::cout &lt;&lt; "not found.\n";

 return 0;
}
</code></pre>

As Piotr said, this will not give you the index, but a yes/no answer. However, is should be the simplest approach, thus the fastest.

blocks|key|2053550|text|您可以使用operator[]像数组一样直接访问向量中的项，而不是使用迭代器从头开始。我猜你已经知道二进制搜索了。在数组中实现它是在任何地方都可以找到的东西，所以我不会在这里向您解释。把向量当作数组来处理。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2053551|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cplusplus.com/reference/vector/vector/operator[]/^0|5|A|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

You can use <a href="http://www.cplusplus.com/reference/vector/vector/operator[]/" rel="nofollow">operator[]</a> to directly access the items in a vector just like an array, instead of using an iterator to start from the begin. I assume you already know Binary Search. Implement it in an array is something you can find anywhere, so I won't explain it to you here. Just treat the vector as an array.

blocks|key|2053596|text|使用std::lower_bound|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2053597|#include+<iterator>
#include+<algorithm>

std::distance(begin(values)
++++++++++++,+std::lower_bound(begin(values),+end(values),+x));|code-block|syntax|javascript|2053598|如果该项不存在，它将为您提供一个比最后一个元素更大的索引。|2053599|演示|2053600|entityMap|0|LINK|mutability|MUTABLE|url|http://en.cppreference.com/w/cpp/algorithm/lower_bound|1|http://coliru.stacked-crooked.com/a/d78cd7574e44703d^0|2|G|2|G|0|0|0|0|0|2|1|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]]|E|$]]|$1|F|3|G|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|15|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|16|8|@]|D|@$9|17|A|18|1|19]]|E|$]]|$1|O|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]|W|$5|R|S|T|E|$U|X]]]]

Use <a href="http://en.cppreference.com/w/cpp/algorithm/lower_bound" rel="nofollow"><code>std::lower_bound</code></a>:

<pre><code>#include &lt;iterator&gt;
#include &lt;algorithm&gt;

std::distance(begin(values)
 , std::lower_bound(begin(values), end(values), x));
</code></pre>

If the item doesn't exist, it'll give you an index one greater than that of the last element.

<a href="http://coliru.stacked-crooked.com/a/d78cd7574e44703d" rel="nofollow">DEMO</a>

blocks|key|892727|text|函数lower_bound可能满足您的需求，您可以如下所示使用它：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|892728|iter+=lower_bound(values.begin(),values.end(),x);|code-block|syntax|javascript|892729|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

The function lower_bound might meet your demand,and you can use it like below:

<pre><code>iter =lower_bound(values.begin(),values.end(),x);
</code></pre>

I have a vector of real values, sorted by increasing order.

These values can be within the range <code>[0, 1]</code>. Then I pick a value <code>x</code> within this range and I need to find which is the index of the smaller value greater or equal to <code>x</code>.

I can solve this problem by iterating over the whole array:

<pre><code>vector&lt;double&gt; values;
double x;

for (auto val : values)
{
 if (x &lt;= values)
 {
 // found
 break;
 }
}
</code></pre>

Is there a faster way to get the same result? I was thinking about a binary search, but how to implement it?

Seek item within vector

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个实数向量，按增加的顺序排序。这些值可以在[0, 1]范围内。然后，我在这个范围内选择一个值x，我需要找到哪个是较小值的索引，它大于或等于x。我可以通过遍历整个数组来解决这个问题：vector<double> values;double x;for (auto val : values){ if (x <=...

问在向量内寻找项目
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