Question

我在Leetcode上遇到了一个名为“ Number of Matching Subsequences”的问题。给您一个字符串S和一个较小字符串的向量，您必须找出多少个较小的字符串是S的子字符串。（不一定是连续的子字符串。）

我以某种方式编写了代码，尽管可以正常工作，但Leetcode上的编译器却超时了。其他人编写的代码与我的几乎相同，但是并没有超时。我想知道是什么让他更快。这是我的：

class Solution {
public:
    int numMatchingSubseq(string S, vector<string>& words) {
        int count = 0;
        vector<set<int>> Svec (26); // keep track of the indices where characters were seen in S
        for (int i = 0; i < S.length(); ++i) Svec[S[i] - 'a'].insert(i);

        for (auto & w : words) { // loop over words and characters within words, finding the soonest the next character appears in S
            bool succeeded = true;
            int current_index = -1;
            for (auto & c : w) {
                set<int> & c_set = Svec[c - 'a'];
                auto it = upper_bound(begin(c_set), end(c_set), current_index);
                if (it == end(c_set)) {
                    succeeded = false;
                    break;
                }
                current_index = *it;
            } // loop over chars
            if (succeeded) count++;
        } //loop over words
        return count;
    }
};

int main() {
    string S = "cbaebabacd";
    vector<string> words {"abc", "abbd", "bbbbd"};
    Solution sol;
    cout << sol.numMatchingSubseq(S, words) << endl;
    return 0;
}

输出

2
Program ended with exit code: 0

他的解决方案不是将索引存储在vector<set<int>>中，而是存储在vector<vector<int>>中。我不明白为什么会有很大的不同。

int numMatchingSubseq (string S, vector<string>& words) {
        vector<vector<int>> alpha (26);
        for (int i = 0; i < S.size (); ++i) alpha[S[i] - 'a'].push_back (i);
        int res = 0;

        for (const auto& word : words) {
            int x = -1;
            bool found = true;

            for (char c : word) {
                auto it = upper_bound (alpha[c - 'a'].begin (), alpha[c - 'a'].end (), x);
                if (it == alpha[c - 'a'].end ()) found = false;
                else x = *it;
            }

            if (found) res++;
        }

        return res;
    }

Answer 1

效率低下：

Blob URL

请参见these std::upper_bound() docs中的注释：

对于非LegacyRandomAccessIterators，迭代器增量的数量是线性的。

您应该使用：

upper_bound(begin(c_set), end(c_set), current_index)

将索引存储在vector <set <int >>与vector <vector <int >>中

1 个答案: