blocks|key|1555927|text|要删除单个元素，可以执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1555928|std::vector<int>+vec;

vec.push_back(6);
vec.push_back(-17);
vec.push_back(12);

//+Deletes+the+second+element+(vec[1])
vec.erase(std::next(vec.begin()));|code-block|syntax|javascript|1555929|或者，一次删除多个元素：|1555930|//+Deletes+the+second+through+third+elements+(vec[1],+vec[2])
vec.erase(std::next(vec.begin(),+1),+std::next(vec.begin(),+3));|1555931|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

To delete a single element, you could do:
<pre><code>std::vector&lt;int&gt; vec;

vec.push_back(6);
vec.push_back(-17);
vec.push_back(12);

// Deletes the second element (vec[1])
vec.erase(std::next(vec.begin()));
</code></pre>
Or, to delete more than one element at once:
<pre><code>// Deletes the second through third elements (vec[1], vec[2])
vec.erase(std::next(vec.begin(), 1), std::next(vec.begin(), 3));
</code></pre>

blocks|key|1340387|text|vector上的erase方法是重载的，因此调用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1340388|vec.erase(vec.begin()+%2B+index);|code-block|syntax|javascript|1340389|当您只想删除单个元素时。|1340390|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

The erase method on std::vector is overloaded, so it's probably clearer to call

<pre><code>vec.erase(vec.begin() + index);
</code></pre>

when you only want to erase a single element.

blocks|key|1340406|text|template+<typename+T>
void+remove(std::vector<T>&+vec,+size_t+pos)
{
++++std::vector<T>::iterator+it+=+vec.begin();
++++std::advance(it,+pos);
++++vec.erase(it);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1340407|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>template &lt;typename T&gt;
void remove(std::vector&lt;T&gt;&amp; vec, size_t pos)
{
 std::vector&lt;T&gt;::iterator it = vec.begin();
 std::advance(it, pos);
 vec.erase(it);
}
</code></pre>

blocks|key|1340794|text|删除带有索引的元素：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1340795|vec.erase(vec.begin()+%2B+index);|code-block|syntax|javascript|1340796|删除具有值的元素：|1340797|vec.erase(find(vec.begin(),vec.end(),value));|1340798|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Erase an element with index :
<pre><code>vec.erase(vec.begin() + index);
</code></pre>
Erase an element with value:
<pre><code>vec.erase(find(vec.begin(),vec.end(),value));
</code></pre>

blocks|key|1340531|text|如果你有一个无序的向量，你可以利用它是无序的，并使用我在CPPCON的Dan+Higgins那里看到的东西|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1340532|template<+typename+TContainer+>
static+bool+EraseFromUnorderedByIndex(+TContainer&+inContainer,+size_t+inIndex+)
{
++++if+(+inIndex+<+inContainer.size()+)
++++{
++++++++if+(+inIndex+!=+inContainer.size()+-+1+)
++++++++++++inContainer[inIndex]+=+inContainer.back();
++++++++inContainer.pop_back();
++++++++return+true;
++++}
++++return+false;
}|code-block|syntax|javascript|1340533|由于列表顺序并不重要，因此只需获取列表中的最后一个元素并将其复制到要删除的项目的顶部，然后弹出并删除最后一个项目。|1340534|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

If you have an unordered vector you can take advantage of the fact that it's unordered and use something I saw from Dan Higgins at CPPCON

<pre><code>template&lt; typename TContainer &gt;
static bool EraseFromUnorderedByIndex( TContainer&amp; inContainer, size_t inIndex )
{
 if ( inIndex &lt; inContainer.size() )
 {
 if ( inIndex != inContainer.size() - 1 )
 inContainer[inIndex] = inContainer.back();
 inContainer.pop_back();
 return true;
 }
 return false;
}
</code></pre>

Since the list order doesn't matter, just take the last element in the list and copy it over the top of the item you want to remove, then pop and delete the last item.

blocks|key|1556009|text|如果您使用大型向量(大小>+100,000)，并且想要删除大量元素，我建议您这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1556010|int+main(int+argc,+char**+argv)+{

++++vector+<int>+vec;
++++vector+<int>+vec2;

++++for+(int+i+=+0;+i+<+20000000;+i%2B%2B){
++++++++vec.push_back(i);}

++++for+(int+i+=+0;+i+<+vec.size();+i%2B%2B)
++++{
++++++++if(vec.at(i)+%253+!=+0)
++++++++++++vec2.push_back(i);
++++}

++++vec+=+vec2;
++++cout+<<+vec.size()+<<+endl;
}|code-block|syntax|javascript|1556011|代码获取vec中不能被3整除的每个数字，并将其复制到vec2。之后，它会在vec中复制vec2。这是相当快的。处理20,000,000个元素，这个算法只需要0.8秒！|1556012|我用erase-method做了同样的事情，它需要很多很多的时间：|1556013|Erase-Version+(10k+elements)++:+0.04+sec
Erase-Version+(100k+elements)+:+0.6++sec
Erase-Version+(1000k+elements):+56+++sec
Erase-Version+(10000k+elements):+...still+calculating+(>30+min)|1556014|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

If you work with large vectors (size > 100,000) and want to delete lots of elements, I would recommend to do something like this:

<pre><code>int main(int argc, char** argv) {

 vector &lt;int&gt; vec;
 vector &lt;int&gt; vec2;

 for (int i = 0; i &lt; 20000000; i++){
 vec.push_back(i);}

 for (int i = 0; i &lt; vec.size(); i++)
 {
 if(vec.at(i) %3 != 0)
 vec2.push_back(i);
 }

 vec = vec2;
 cout &lt;&lt; vec.size() &lt;&lt; endl;
}
</code></pre>

The code takes every number in vec that can't be divided by 3 and copies it to vec2. Afterwards it copies vec2 in vec. It is pretty fast. To process 20,000,000 elements this algorithm only takes 0.8 sec!

I did the same thing with the erase-method, and it takes lots and lots of time:

<pre><code>Erase-Version (10k elements) : 0.04 sec
Erase-Version (100k elements) : 0.6 sec
Erase-Version (1000k elements): 56 sec
Erase-Version (10000k elements): ...still calculating (&gt;30 min)
</code></pre>

blocks|key|1340750|text|我建议你读一下这篇文章，因为我相信这就是你想要的。https://en.wikipedia.org/wiki/Erase%25E2%2580%2593remove_idiom|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1340751|如果您使用例如|1340752|+vec.erase(vec.begin()+%2B+1,+vec.begin()+%2B+3);|code-block|syntax|javascript|1340753|您将擦除向量的n个-th元素，但当您擦除第二个元素时，向量的所有其他元素都将移位，向量大小将为-1。如果循环遍历向量，这可能会有问题，因为向量大小()正在减小。如果你有这样的问题，提供的链接建议使用标准C%2B%2B库中的现有算法。和"remove“或"remove_if”。|1340754|希望这对你有所帮助|1340755|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Erase%25E2%2580%2593remove_idiom^0|P|1K|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|6|7|10|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|11|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|C|$]]|$1|M|3|N|5|6|7|13|8|@]|9|@]|C|$]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]]]

I suggest to read this since I believe that is what are you looking for.<a href="https://en.wikipedia.org/wiki/Erase%E2%80%93remove_idiom" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Erase%E2%80%93remove_idiom</a>

If you use for example 

<pre><code> vec.erase(vec.begin() + 1, vec.begin() + 3);
</code></pre>

you will erase n -th element of vector but when you erase second element, all other elements of vector will be shifted and vector sized will be -1. This can be problem if you loop through vector since vector size() is decreasing. If you have problem like this provided link suggested to use existing algorithm in standard C++ library. and "remove" or "remove_if". 

Hope that this helped

blocks|key|1556406|text|如果需要擦除for循环中的元素，请执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1556407|for(int+i+=+0;+i+<+vec.size();+i%2B%2B){

++++if(condition)
++++++++vec.erase(vec.begin()+%2B+i);

}|code-block|syntax|javascript|1556408|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|$]]

if you need to erase an element inside of a for-loop, do the following:
<pre><code>for(int i = 0; i &lt; vec.size(); i++){

 if(condition)
 vec.erase(vec.begin() + i);

}
</code></pre>

blocks|key|1556175|text|这是另一种方法，如果你想删除一个元素，通过在向量中找到它的值，你只需要对向量做这件事。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1556176|vector<int>+ar(n);
ar.erase(remove(ar.begin(),+ar.end()),+(place+your+value+here+from+vector+array));|code-block|syntax|javascript|1556177|它会将你的价值从这里移走。谢谢|1556178|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

here is one more way to do this if you want to delete a element by finding this with its value in vector,you just need to do this on vector.

<pre><code>vector&lt;int&gt; ar(n);
ar.erase(remove(ar.begin(), ar.end()), (place your value here from vector array));
</code></pre>

it will remove your value from here.
thanks

blocks|key|1340666|text|这个怎么样？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1340667|void+squeeze(vector<int>+&v)
{
++++int+j+=+0;
++++for+(int+i+=+1;+i+<+v.size();+i%2B%2B)
++++++++if+(v[i]+!=+v[j]+&&+%2B%2Bj+!=+i)
++++++++++++v[j]+=+v[i];
++++v.resize(j+%2B+1);
}|code-block|syntax|javascript|1340668|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|$]]

How about this?

<pre><code>void squeeze(vector&lt;int&gt; &amp;v)
{
 int j = 0;
 for (int i = 1; i &lt; v.size(); i++)
 if (v[i] != v[j] &amp;&amp; ++j != i)
 v[j] = v[i];
 v.resize(j + 1);
}
</code></pre>

blocks|key|1340635|text|最快的方法(通过时间复杂度()=常量来编程竞赛)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1340636|可在1秒内擦除100M项；|1340637|++++vector<int>+it+=+(vector<int>::iterator)+&vec[pos];
++++vec.erase(it);|code-block|syntax|javascript|1340638|最具可读性的方式：vec.erase(vec.begin()+%2B+pos);|offset|length|style|CODE|1340639|entityMap^0|0|0|0|9|T|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@$K|U|L|V|M|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

the fastest way (for programming contests by time complexity() = constant) 

can erase 100M item in 1 second;

<pre><code> vector&lt;int&gt; it = (vector&lt;int&gt;::iterator) &amp;vec[pos];
 vec.erase(it);

</code></pre>

and most readable way :
<code>vec.erase(vec.begin() + pos);
</code>

I have a std::vector&lt;int>, and I want to delete the n'th element. How do I do that?

<pre><code>std::vector&lt;int&gt; vec;

vec.push_back(6);
vec.push_back(-17);
vec.push_back(12);

vec.erase(???);
</code></pre>

How do I erase an element from std::vector<> by index?

我有一个std::vector，我想删除第n个元素。我该怎么做？std::vector<int> vec;vec.push_back(6);vec.push_back(-17);vec.push_back(12);vec.erase(???);

问如何通过索引从std::vector<>中擦除元素？
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问如何通过索引从std::vector<>中擦除元素？EN

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问如何通过索引从std::vector<>中擦除元素？
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