C++ 如何检查/查找项目是否在 DEQUE 中
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/8930417/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to check/find if an item is in a DEQUE
提问by george mano
In the code above the else-if part gives me error. The meaning of else-if is: else if the value of x isn't in the deque then...
在上面的代码中，else-if 部分给了我错误。else-if 的意思是：else 如果 x 的值不在双端队列中，那么...
#include <iostream>
#include <ctime>
#include <stack>
#include <deque>
#include <algorithm>
deque<char> visited;
char x;
if (x==target[4][4])
{
visited.push_back(x);
return (visited);
}
else if (!(find(visited.begin(), visited.end(), x)))
{
visited.push_back(x);
}
ERROR:no operator "!" matches these operands
错误：没有运算符“！” 匹配这些操作数
回答by kennytm
If std::findcannot find the specific value, it will return the "end" of the iterator pair.
如果std::find找不到特定值，它将返回迭代器对的“结束”。
else if (std::find(visited.begin(), visited.end(), x) == visited.end())
{
// process the case where 'x' _is_not_ found between
// visited.begin() and visited.end()
Edit: If you want to know if xisin the deque, just reverse the condition.
编辑：如果您想知道x是否在双端队列中，只需反转条件即可。
else if (std::find(visited.begin(), visited.end(), x) != visited.end())
{
// process the case where 'x' _is_ found between
// visited.begin() and visited.end()
Edit: If you are unfamiliar with the iterator concept in C++, please read Understanding Iterators in the STL.
编辑：如果您不熟悉 C++ 中的迭代器概念，请阅读了解 STL 中的迭代器。
回答by Hridaynath
For those who visited this page to simply know how to check/find elements in dequeue. A quick solution is as below:
对于那些访问此页面的人来说，只需知道如何检查/查找出列中的元素。一个快速的解决方案如下：
Use std::find()method:
使用std::find()方法：
numbers.push_back(10);
numbers.push_front(20);
numbers.push_back(30);
numbers.push_front(40);
deque<int>::iterator it = find(numbers.begin(), numbers.end(), 20);
if(it!=numbers.end())
{
// Do your stuff. Here I am simply deleting the element
it = numbers.erase(it);
// Note: Always save returned iterator from erase/insert method, otherwise
// iterator will point to deleted resource, which leads to undefined behaviour.
}
Hope this will help somebody. :)
希望这会帮助某人。:)
