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Codility-PassingCars

2024-09-15 01:27:53   215人阅读

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:

class Solution { public int solution(int[] 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.

　　思路： 直觉 冒泡数组排列 0 1 组合 -> O(n2)   X

               O(n) 要求只能遍历一遍数组。而且只允许（0,1），不允许（1,0）。。。。。囧么办呢？

              （0，1）组合。☆1可以和前面所有的0组合（由前遍历）或者 0可以和后面所有的1组合（由后遍历）。

 

class Solution {
    public int solution(int[] A) {
        // write your code in Java SE 
        int N = A.length;
        int sum_1 = 0;
        int result = 0;
 
        for(int i=N-1;i>=0; i--){
            if(A[i]==1)
                sum_1 = sum_1+1 ;
            else
                result = result + sum_1 ;
            if(result>1000000000) return -1;
        }
        return result;
    }
}

 

Codility-PassingCars


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