JavaScript-Simple Baidu School recruit questions, see a variety of solutions?

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I have not done this problem, the provisional thought of the algorithm:

Two points to the target, assuming that the current number is num , then traverse each row, for the first i row, no greater than num the number of numbers is min(num / i, m) , accumulated after the total count cnt .

If cnt less than k then to the right half of the range continue to find, otherwise to the left half to continue to find.

Time complexity, more than O(n * log(n * m)) sufficient.

Direct use of subscript query, each row is a multiple of the first row

It can be found that the multiplication table results in the size of the diagonal line, so follow this method to find the law.

N*m Matrix results you need it.
I'll give you a way to build a 0-n*m array A, each element is 0,
Then you have two for loops to calculate each number of T in the Matrix, will a[t]++,
After the calculation you have a one-dimensional array that might be "0,1,2,3,1,0,2,5, ... 】
You add from left to right until the result is greater than or equal to K, and the subscript i is the result you want

def multiply(n, m, k):    if k > n * m:        return None    l = []    for i in range(1, n + 1):        for j in range(1, m + 1):            l.append(i * j)    return l[k]print multiply(2,3, 7)

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