Codility-PassingCars

Posted 2020-09-11

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;
    }
}