C++ 9 lines hash table easy to understand

  • 4
        vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
            int min = INT_MAX;
            for(int i = 0; i < list1.size(); i++) m[list1[i]] = i;
            for(int i = 0; i < list2.size(); i++)
                if(m.count(list2[i]) != 0)
                    if(m[list2[i]] + i < min) min = m[list2[i]] + i, res.clear(), res.push_back(list2[i]);
                    else if(m[list2[i]] + i == min) res.push_back(list2[i]);
            return res;

  • 0

    brilliant! with perfect run time.But you erase and reload the res everytime you find a smaller one,is there any other way to do this?

  • 1

    @horanweihaoran As you mentioned, clear() operation happens only when we found a smaller one, that is

    1. Two lists share a common interest.
    2. The index sum is smaller than previous one.

    Combined 1 and 2, although clear() is a linear time operation, we can see that the clear() operation will only occur a few times, and in average, it only costs a constant time, that's why the runtime is decent.

    (BTW, clear() is linear, but actually we don't have much data in res (right?), which makes res.clear() itself an O(1) operation.)

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