• 【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:

     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 }
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  • 原文地址:https://www.cnblogs.com/wei-li/p/PassingCars.html
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