问题:

def solution(M, A):
    result = [0] * M
    maxCount = 0
    setAll = 0

    for i in range(0,len(A)):
        if (A[i] == M + 1):
            setAll += maxCount
            maxCount = 0
            result = [0] * M
        else:
            result[A[i] - 1] += 1

            if (result[A[i] - 1] > maxCount):
                maxCount = result[A[i] - 1]

    for j in range(0,len(result)):
        result[j] += setAll

    return result


A = [ 1, 1, 1, 1, 2, 3] 
M = 2

print solution(M, A) # result = [ 4, 4 ]


A = [ 1, 2, 2, 4, 1, 1] 
M = 3

print solution(M, A) # result = [ 4, 2, 2 ]

By my count, solution() loops through A N times and then loops result M times thus N+M. However, an online test said that it was instead N*M, leaving me stumped.

回答1:

It is O(M + N); there are no nested loops here. This can be reduced to the cost for the larger number; asymptotically, the smaller loop is not going to matter, making it O(N).

First you loop over A elements (N iterations), then separately you loop over M elements.

回答2:

It is O(N), because the there are N + M loops for a given input N, M. That reduces to the larger of the two (let's say that is N) for the purposes of time complexity, because we only take the most significant term. It would be O(N*M) if the second loop was nested inside the first.