Question

以下描述算法的名称是什么？

描述：

如何在给定模式中测试序列是否符合另一个序列？

例如：

The pattern: the number appears in the same order.


bool isOrderValid(vector<int>& mainVec, vector<int>& subVec) {
    // how to implement this function?
}

test#1: isOrderValid({1, 2, 3, 4}, {2, 4}); // should return true
test#2: isOrderValid({1, 2, 3, 4}, {4, 2}); // should return false

说明：

测试＃1：子序列为2,4;在主序列中，2出现在4之前，所以顺序正确。

测试＃2：子序列是4,2;但是，在主序列中，4出现在2之后，因此顺序不正确。

注意：两个数组中可能都有重复的条目。例如：

isOrderValid({3, 6, 3, 1, 2, 3}, {3, 1, 3}); // should return true

Answer 1

这可以非常简单地实现（我在这里使用函数对象）：

class base_pattern_matcher{
    public:
        virtual bool operator()(const vector<int>& a , const vector<int>& b) = 0;
}

class in_order_pattern_matcher : public base_pattern_matcher{
    public:
        bool operator()(const vector<int>& a , const vector<int>& b){
            vector<int>::iterator iter_a = a.begin() , iter_b = b.begin();

            while(iter_b != b.end() && iter_a != a.end()){
                if(*iter_b == *iter_a)
                    //we found an occurence in a that matches the searched element in b
                    //--> search for the next element
                    iter_b++;

                 //check if the next element matches the searched element
                 iter_a++;
            }

            //we have found all elements of b in the given order
            return iter_b == b.end();
        };
}

isOrderValid(const vector<int>& a , const vector<int>& b , const base_pattern_matcher& matcher){
    return matcher(a , b);
}

Answer 2

您可以为此编写标准库样式算法。此示例采用两个迭代器对并返回bool：

#include <iostream>
#include <vector>

template <typename InIt1, typename InIt2>
bool is_subsequence(InIt1 first1, InIt1 last1, InIt2 first2, InIt2 last2)
{
    if (first2 == last2) {
        return false; // sub empty (should this return true?)
    }
    for (; first1 != last1; ++first1) {
        if (*first1 == *first2) {
            if (++first2 == last2) {
                return true; // sub exhausted
            }
        }
    }
    return false; // seq exhausted
}

int main()
{
    std::vector<int> a = { 1, 2, 3, 4 }, b = { 2, 4 }, c = { 4, 2 };
    std::vector<int> d = { 3, 6, 3, 1, 2, 3 }, e = { 3, 1, 3 };

    std::cout << is_subsequence(a.begin(), a.end(), b.begin(), b.end()) << '\n';
    std::cout << is_subsequence(a.begin(), a.end(), c.begin(), c.end()) << '\n';
    std::cout << is_subsequence(d.begin(), d.end(), e.begin(), e.end()) << '\n';
}

Executed on ideone.com

如果您喜欢编写通用函数，可以使其更灵活，更易于使用：

#include <iostream>
#include <list>
#include <vector>

template <typename InIt1, typename InIt2, typename Compare = std::equal_to<>>
bool is_subsequence(InIt1 first1, InIt1 last1, InIt2 first2, InIt2 last2, Compare cmp = Compare{})
{
    if (first2 == last2) {
        return false; // sub empty (should this return true?)
    }
    for (; first1 != last1; ++first1) {
        if (cmp(*first1, *first2)) {
            if (++first2 == last2) {
                return true; // sub exhausted
            }
        }
    }
    return false; // seq exhausted
}

template <typename Seq, typename Sub, typename Compare = std::equal_to<>>
bool is_subsequence(const Seq &seq, const Sub &sub, Compare cmp = Compare{})
{
    return is_subsequence(std::begin(seq), std::end(seq), std::begin(sub), std::end(sub), cmp);
}

int main()
{
    std::vector<int> a = { 1, 2, 3, 4 }, b = { 2, 4 };
    std::list<int> c = { 4, 2 };
    std::vector<int> d = { 3, 6, 3, 1, 2, 3 };
    int e[] = { 3, 1, 3 };

    std::cout << is_subsequence(a, b) << '\n';
    std::cout << is_subsequence(a, b, [](int lhs, int rhs) { return lhs == rhs; }) << '\n';
    std::cout << is_subsequence(a, c) << '\n';
    std::cout << is_subsequence(d, e) << '\n';
}

Executed on ideone.com

如何测试序列是否符合给定模式中的另一个序列？

2 个答案: