• Codility----PassingCars


    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.

    Solution
     
    Programming language used: Java
    Code: 02:08:07 UTC, java, final, score:  100
    // you can also use imports, for example:
    // import java.util.*;
    
    // you can write to stdout for debugging purposes, e.g.
    // System.out.println("this is a debug message");
    
    class Solution {
        public int solution(int[] A) {
            // write your code in Java SE 8
            int zeroCnt=0, oneCnt =0;
            for(int i=0; i<A.length; i++) {
                if(A[i] == 0) {
                    zeroCnt += 1;
                } else if(A[i] == 1) {
                    oneCnt += zeroCnt;
                }
                if(oneCnt > 1000000000)
                return -1;
            }        
            return oneCnt;
        }
    }



    https://codility.com/demo/results/training8U5YGT-RJR/
  • 相关阅读:
    Longest Valid Parentheses
    [转载]ios入门篇 -hello Word(1)
    EXTJS 4 动态grid
    Spring AOP JPA
    Jchart 演示
    HSQLDB JPA GeneratedValue
    Antlr 练习
    回火方程
    URL decode 解决中文目录的乱码问题
    Arduino IIC lcd1602
  • 原文地址:https://www.cnblogs.com/samo/p/6780348.html
Copyright © 2020-2023  润新知