HackerRank Js-Week4 3. Number Line Jumps

    AllenLi
    HackerRank

    Complete the function kangaroo in the editor below.
    kangaroo has the following parameter(s):
    int x1, int v1: starting position and jump distance for kangaroo 1
    int x2, int v2: starting position and jump distance for kangaroo 2

    在下面的編輯器中完成袋鼠功能。
袋鼠有以下參數：
int x1, int v1: 袋鼠1的起始位置和跳躍距離
int x2、int v2：袋鼠2的起始位置和跳躍距離

    Example 1:

    x1 = 2
v1 = 1
x2 = 1
v2 = 2

After one jump, they are both at x = 3, (x1 + v1 = 2 + 1, x2 + v2 = 1 + 2), so the answer is YES.

    solution:
    首先我們先了解 x1, v1, x2, v2 分別代表兩隻袋鼠的起始位置與速度。
    接著我們計算兩隻袋鼠的相遇時間(meet = (x2 – x1) / (v1 – v2))，
    並回傳是否相遇。

    Code 1: BigO(1)

    'use strict';

const fs = require('fs');

process.stdin.resume();
process.stdin.setEncoding('utf-8');

let inputString = '';
let currentLine = 0;

process.stdin.on('data', function(inputStdin) {
    inputString += inputStdin;
});

process.stdin.on('end', function() {
    inputString = inputString.split('\n');

    main();
});

function readLine() {
    return inputString[currentLine++];
}

/*
 * Complete the 'kangaroo' function below.
 *
 * The function is expected to return a STRING.
 * The function accepts following parameters:
 *  1. INTEGER x1
 *  2. INTEGER v1
 *  3. INTEGER x2
 *  4. INTEGER v2
 */

function kangaroo(x1, v1, x2, v2) {
    // Write your code here
    let meet = (x2 - x1) / (v1 - v2);

    if (meet === Math.round(meet) && meet > 0 && meet !== Infinity) {
        return "YES";
    } 

    return 'NO';
}

function main() {
    const ws = fs.createWriteStream(process.env.OUTPUT_PATH);

    const firstMultipleInput = readLine().replace(/\s+$/g, '').split(' ');

    const x1 = parseInt(firstMultipleInput[0], 10);

    const v1 = parseInt(firstMultipleInput[1], 10);

    const x2 = parseInt(firstMultipleInput[2], 10);

    const v2 = parseInt(firstMultipleInput[3], 10);

    const result = kangaroo(x1, v1, x2, v2);
    ws.write(result + '\n');

    ws.end();
}

    FlowChart:
    Example 1

    x1 = 2
v1 = 1
x2 = 1
v2 = 2

meet = -1 / -1 //1

meet === Math.round(meet) //meet = 1 === Math.round(meet)
meet > 0 //meet = 1 > 0
meet !== Infinity //不等於無窮大

return "YES";

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