HackerRank Js-Week2 1. Lonely Integer

    HackerRank

    Given an array of integers, where all elements but one occur twice, find the unique element.

    給予一個整數的陣列,其中除了一個元素以外的所有元素都出現兩次,找出那個唯一的元素。
    

    Example 1:

    a = [1, 2, 3, 4, 3, 2, 1]
    
    The unique element is 4.
    

    solution:
    先宣告一個空的物件叫做 hashTable,並將題目給的 a 當作 key 放入,
    接著將 key 出現的次數記錄到物件中key 對應的 value,
    最後判斷 value 為 1 的 key 進行回傳。

    Code 1: BigO(2n)

    '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 'lonelyinteger' function below.
     *
     * The function is expected to return an INTEGER.
     * The function accepts INTEGER_ARRAY a as parameter.
     */
    
    function lonelyinteger(a) {
        // Write your code here    
        let hashTable = {}
        
        for (let i = 0; i < a.length; i++) {
            hashTable[a[i]] = (hashTable[a[i]] || 0) + 1
        }
        for (let value in hashTable) {
            if (hashTable[value] === 1) return value;
        }
    }
    
    function main() {
        const ws = fs.createWriteStream(process.env.OUTPUT_PATH);
    
        const n = parseInt(readLine().trim(), 10);
    
        const a = readLine().replace(/\s+$/g, '').split(' ').map(aTemp => parseInt(aTemp, 10));
    
        const result = lonelyinteger(a);
    
        ws.write(result + '\n');
    
        ws.end();
    }
    

    FlowChart:
    Example 1

    a = [1, 2, 3, 4, 3, 2, 1]
    
    hashTable = {
      '1' : 2,
      '2' : 2,
      '3' : 2,
      '4' : 1
    }
    
    hashTable[4] = 1 === 1
    return 4;