codility: PassingCars

最新推荐文章于 2025-09-12 20:46:57 发布

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最新推荐文章于 2025-09-12 20:46:57 发布

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文章标签：大数据 c/c++

原文链接：http://www.cnblogs.com/parapax/p/3635236.html

Task description

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:

int solution(vector<int> &A);

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

The function should return −1 if the number 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 within the range [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.

从后往前遍历就好，但是交了几遍居然都过不了大数据，在expected -1的时候我还返回了实际的res，于是我就搜了下别人的代码，发现总有类似res>1000000000这么一句，我还很纳闷这个数哪里来的，直到重新看了一遍题目，才发现了The function should return −1 if the number of passing cars exceeds 1,000,000,000.

简直对自己无语。

醒醒吧亲！看清楚题目再写好吗！

希望自己别再因为粗心而摔在原本可以搞定的事情上，以此为记。

// you can also use includes, for example:
// #include <algorithm>
int solution(vector<int> &A) {
    // write your code in C++98
    int countOf1 =0;
    int res = 0;
    for(int i = A.size()-1;i>=0;i--) {
        if(A[i]==1)
            countOf1+=1;
        else {
            res+=countOf1;
            if(res>1000000000)
                return -1;
        }
    }
    return res;
}