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：
1 int solution(vector<int> &A) { 2 int res = 0; 3 int n = A.size(); 4 int last = 0; 5 for(int i = n-1; i >= 0; i--){ 6 if(A[i] == 0) continue; 7 A[i] += last; 8 last = A[i]; 9 } 10 last = 0; 11 for(int i = n-1; i >= 0; i--){ 12 if(A[i] == 0) 13 res += last; 14 if(A[i] != 0) 15 last = A[i]; 16 } 17 return res; 18 }