您可以在递归lambda中捕获引用吗?

时间:2019-04-25 13:58:53

标签: c++ c++11 recursion lambda

我已经使用带有备注的递归解决方案完全解决了HackerRank(https://www.hackerrank.com/challenges/ctci-recursive-staircase/problem)上的一个特定问题:

std::map<int, int> memoize;

int davis_staircase(int n) {
    if (n == 0) {
        return 1;
    } else if (n < 0) {
        return 0;
    }

    auto find = memoize.find(n);
    if (find != memoize.end()) {
        return find->second;
    }

    int num_steps = davis_staircase(n - 1) + davis_staircase(n - 2) + davis_staircase(n - 3);
    memoize[n] = num_steps;

    return num_steps;
}

我想隐藏用作查询的全局std::map(不使用类),并认为我会尝试创建可以递归调用的lambda并捕获缓存/映射引用。我尝试了以下方法:

int davis_staircase_2(int n) {

    std::map<int, int> memo;

    //auto recurse = [&memo](int n) -> int {                    // attempt (1)
    //std::function<int(int)> recurse = [&memo](int n) -> int { // attempt (2)
    std::function<int(std::map<int, int>&, int)> recurse = [](std::map<int, int>& memo, int n) -> int { // attempt (3)
        if (n == 0) {
            return 1;
        } else if (n < 0) {
            return 0;
        }

        auto find = memo.find(n);
        if (find != memo.end()) {
            return find->second;
        }

        //int num_steps = recurse(n - 1) + recurse(n - 2) + recurse(n - 3); // attempt (1) or (2)
        int num_steps = recurse(memo, n - 1) + recurse(memo, n - 2) + recurse(memo, n - 3); // attempt (3)

        memo[n] = num_steps;

        return num_steps;
    };

    //return recurse(n); // attempt (1) or (2)
    return recurse(memo, n); // attempt (3)
}

我在上面交错了3次稍有不同的尝试,但是我无法编译。是我想做的事吗?

我在MacOS上使用clang:

Apple LLVM version 10.0.0 (clang-1000.10.44.4)
Target: x86_64-apple-darwin18.2.0
Thread model: posix

3 个答案:

答案 0 :(得分:3)

您忘记捕获recurse,因此您的代码可能是

std::function<int(int)> recurse = [&recurse, &memo](int n) -> int { // attempt (2)

std::function<int(int)> recurse = [&](int n) -> int { // attempt (2)

对于// attempt (3),以相同的方式:

std::function<int(std::map<int, int>&, int)> recurse = [&recurse](std::map<int, int>& memo, int n) -> int { // attempt (3)

// attempt (1)无法原样固定,因为在定义之前使用了recurse类型。

要在没有std::function的情况下执行此操作,可以使用Y-combinator(对于通用lambda,需要C ++ 14):

int davis_staircase_2(int n) {
    std::map<int, int> memo;
    auto recurse = [&memo](auto self, int n) -> int { // attempt (4)
        if (n == 0) {
            return 1;
        } else if (n < 0) {
            return 0;
        }

        auto find = memo.find(n);
        if (find != memo.end()) {
            return find->second;
        }

        int num_steps = self(self, n - 1) + self(self, n - 2) + self(self, n - 3); // attempt (4)

        memo[n] = num_steps;

        return num_steps;
    };
    return recurse(recurse, n); // attempt (4)
}

答案 1 :(得分:0)

您不需要递归函数...

int stepPerms(int n) {

  std::map<int, int> memoize;

  memoize[-2] = 0;
  memoize[-1] = 0;
  memoize[0] = 1;

 for(int i=1;i<=n;++i)
 {
   memoize[i] = memoize[i - 1] + memoize[i - 2] + memoize[i-3];
 }

 return memoize[n];
}

答案 2 :(得分:0)

您可以执行递归lambda,而无需类型擦除(std :: function)。这是使用通用lambda来完成的:

auto recurse = [](auto lambda) {
    return [lambda](auto&&... args) {
        return lambda(lambda, std::forward<decltype(args)>(args)...);
    };
};

auto my_recursive_lambda = recurse([](auto self, std::map<int, int>& memo, int n) {
    if (n == 0) {
        return 1;
    } else if (n < 0) {
        return 0;
    }

    auto find = memo.find(n);
    if (find != memo.end()) {
        return find->second;
    }

    int num_steps = self(self, memo, n - 1) + self(self, memo, n - 2) + self(self, memo, n - 3);

    memo[n] = num_steps;

    return num_steps;
});

my_recursive_lambda(memo, n); // magic!

如果您确实需要c ++ 11,则需要std::function

auto recurse = std::function<int(std::map<int, int>&, int)>{};

recurse = [&recurse](std::map<int, int>& memo, int n) {
    // same as you tried.
}

或者,如果您放弃方便,可以手动滚动lambda类型:

struct {
    auto operator()(std::map<int, int>& memo, int n) -> int {
        auto&& recurse = *this;
        if (n == 0) {
            return 1;
        } else if (n < 0) {
            return 0;
        }

        auto find = memo.find(n);
        if (find != memo.end()) {
            return find->second;
        }

        //int num_steps = recurse(n - 1) + recurse(n - 2) + recurse(n - 3); // attempt (1) or (2)
        int num_steps = recurse(memo, n - 1) + recurse(memo, n - 2) + recurse(memo, n - 3); // attempt (3)

        memo[n] = num_steps;

        return num_steps;
    }
} recurse{};