【题解】【数组】【Prefix Sums】【Codility】Passing Cars

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.

思路：

这题用Prefix Sums做非常简单，不过题目要求“return −1 if the number of passing cars exceeds 1,000,000,000.”这点我忘了处理，所以大数据没过得了66分，各位看官自行脑补一下。

代码：

 int solution(vector<int> &A) {
     int res = ;
     int n = A.size();
     int last = ;
     for(int i = n-; i >= ; i--){
         if(A[i] == ) continue;
         A[i] += last;
         last = A[i];
     }
     last = ;
     for(int i = n-; i >= ; i--){
         if(A[i] == )
             res += last;
         if(A[i] != )
             last = A[i];
     }
     return res;
 }