Question

我目前正在练习编程问题并且没有兴趣，我正在Lua尝试一些Codility练习。我一直坚持Passing Cars问题。

问题：

A non-empty zero-indexed array A consisting of N integers is given. The consecutive elements of array A represent consecutive cars on a road.

Array A contains only 0s and/or 1s:

0 represents a car traveling east,
1 represents a car traveling west.
The goal is to count passing cars. We say that a pair of cars (P, Q), where 0 ≤ P < Q < N, is passing when P is traveling to the east and Q is traveling to the west.

For example, consider array A such that:

  A[0] = 0
  A[1] = 1
  A[2] = 0
  A[3] = 1
  A[4] = 1
We have five pairs of passing cars: (0, 1), (0, 3), (0, 4), (2, 3), (2, 4).

Write a function:

function solution(A)

that, given a non-empty zero-indexed array A of N integers, returns the number of pairs of passing cars.

The function should return −1 if the number of pairs of passing cars exceeds 1,000,000,000.

For example, given:

  A[0] = 0
  A[1] = 1
  A[2] = 0
  A[3] = 1
  A[4] = 1
the function should return 5, as explained above.

Assume that:

N is an integer within the range [1..100,000];
each element of array A is an integer that can have one of the following values: 0, 1.
Complexity:

expected worst-case time complexity is O(N);
expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).
Elements of input arrays can be modified.

我在Lua的尝试一直在失败，但我似乎无法找到问题。

local function solution(A)
    local zeroes = 0
    local pairs = 0
    for i = 1, #A do
        if A[i] == 0 then
           zeroes = zeroes + 1
        else
            pairs = pairs + zeroes
            if pairs > 1e9 then
                return -1
            end
        end
    end
    return pairs
end

就时空复杂性约束而言，我认为它应该通过，所以我似乎无法找到问题。我究竟做错了什么？任何建议或提示，以使我的代码更有效将不胜感激。仅供参考：当所需的示例结果为5时，我会得到2的结果。

Answer 1

问题陈述说A是基于0的，所以如果我们忽略第一个并从1开始，输出将是2而不是5.在Lua中应避免使用基于0的表，它们违反惯例并将导致很多失误的错误：for i=1,#A do不会做你想做的事。

function solution1based(A)
    local zeroes = 0
    local pairs = 0
    for i = 1, #A do
        if A[i] == 0 then
           zeroes = zeroes + 1
        else
            pairs = pairs + zeroes
            if pairs > 1e9 then
               return -1
            end
        end
    end
    return pairs
end
print(solution1based{0, 1, 0, 1, 1}) -- prints 5 as you wanted

function solution0based(A)
    local zeroes = 0
    local pairs = 0
    for i = 0, #A do
        if A[i] == 0 then
           zeroes = zeroes + 1
        else
            pairs = pairs + zeroes
            if pairs > 1e9 then
               return -1
            end
        end
    end
    return pairs
end

print(solution0based{[0]=0, [1]=1, [2]=0, [3]=1, [4]=1}) -- prints 5

Codility：在Lua经过汽车

1 个答案: