使用STL算法查找集合中的前两个非相邻元素 [英] Find First Two Non-Adjacent Elements in a set Using an STL Algorithm

问题描述

所以我真的为此感到挣扎，即使现在我对自己的解决方案也不满意.

So I really struggled with this and even now I am not happy with my solution.

我有一个set，至少包含0，并且可能包含其他正数int.我需要在set中找到第一个正数 not .

I have a set that at least contains 0, and may contain other positive ints. I need to find the first positive number not in the set.

因此，编写标准的while循环来完成此操作很容易.

So writing a standard while-loop to accomplish this is easy.

i = foo.begin();
while (i != prev(foo.end()) && *i + 1 == *next(i)){
    ++i;
}
cout << "First number that cannot be formed " << *i + 1 << endl;

但是当我尝试编写该循环的STL算法版本时，会得到类似该循环的错误消息:

But when I try to write an STL algorithm version of the loop I get something that fails like this loop:

auto i = foo.begin();
while (++i != prev(foo.end()) && *prev(i) + 1 == *i);
cout << "First number that cannot be formed " << *prev(i) + 1 << endl;

在以下情况下，这两个循环均正确产生 3 :

Both of these loops correctly yield 3 in the case of:

set<int> foo{0, 1, 2, 4};

但是在这种情况下，第二个循环错误地产生了3而不是 4 :

But the second loop incorrectly yields 3 instead of 4 in this case:

set<int> foo{0, 1, 2, 3};

如何使用STL算法编写此代码并完成第一个循环的行为?

看到一些答案后，我想增加难度.我真正想要的是不需要临时变量并且可以放在cout语句中的东西.

After seeing some of the answers, I'd like to increase the difficulty. What I really want is something that doesn't require temporary variables and can be placed in a cout statement.

推荐答案

循环的问题是您过早停止了一个元素.这有效:

The problem with your loop is that you stop one element too early. This works:

while (++i != foo.end() && *prev(i) + 1 == *i);

与第一个循环的区别是条件*prev(i) + 1 == *i)而不是*i + 1 == *next(i)；您检查的范围必须相应地移动.

The difference to the first loop is the condition *prev(i) + 1 == *i) instead of *i + 1 == *next(i); the range you check has to shift accordingly.

您也可以使用std::adjacent_find:

auto i = std::adjacent_find(begin(s), end(s), [](int i, int j) { return i + 1 != j; });

if(i == s.end()) {
  std::cout << *s.rbegin() + 1 << std::endl;
} else {
  std::cout << *i + 1 << std::endl;
}

对编辑的响应:使其完全内联的一种方法是

Response to the edit: A way to make it prettily inlineable is

  std::cout << std::accumulate(begin(s), end(s), 0,
                               [](int last, int x) {
                                 return last + 1 == x ? x : last;
                               }) + 1 << '\n';

...但是效率较低，因为它在发现间隙时不会短路.造成短路的另一种方法是

...but this is less efficient because it does not short-circuit when it finds a gap. Another way that does short-circuit is

std::cout << *std::mismatch(begin     (s),
                            prev(end  (s)),
                            next(begin(s)),
                            [](int lhs, int rhs) { 
                              return lhs + 1 == rhs;
                            }).first + 1 << '\n';

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