projecteuler Problem 8 Largest product in a series

The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?

譯文：

觀察著1000個數字，通過運算可得，最大的連續的四位數的乘機為5832，求出最大的連續的13位數的乘機？

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第一次code:

import java.util.ArrayList;
import java.util.List;

public class Main
{
    public static void main(String[] args)   
    {
        System.out.println(num(run()));
    }
    /*
    *  將1000個數放到一個字符串
    *  將字符串轉換成字符串數組
    *  計算每連續的13位數的乘機，並放置在Arraylist數組中
    *
    *  注意：乘機的類型為long，如果為int的話，數據結果會超出，導致計算結果出錯
    */
    public static List run()
    {
        String a="73167176531330624919225119674426574742355349194934"
                +"96983520312774506326239578318016984801869478851843"
                +"85861560789112949495459501737958331952853208805511"
                +"12540698747158523863050715693290963295227443043557"
                +"66896648950445244523161731856403098711121722383113"
                +"62229893423380308135336276614282806444486645238749"
                +"30358907296290491560440772390713810515859307960866"
                +"70172427121883998797908792274921901699720888093776"
                +"65727333001053367881220235421809751254540594752243"
                +"52584907711670556013604839586446706324415722155397"
                +"53697817977846174064955149290862569321978468622482"
                +"83972241375657056057490261407972968652414535100474"
                +"82166370484403199890008895243450658541227588666881"
                +"16427171479924442928230863465674813919123162824586"
                +"17866458359124566529476545682848912883142607690042"
                +"24219022671055626321111109370544217506941658960408"
                +"07198403850962455444362981230987879927244284909188"
                +"84580156166097919133875499200524063689912560717606"
                +"05886116467109405077541002256983155200055935729725"
                +"71636269561882670428252483600823257530420752963450";
        char [] b = a.toCharArray();
        List<Long>list = new ArrayList<Long>();
        for(int i=0;i<b.length-13;i++)
        {
            list.add(Long.valueOf(String.valueOf(b[i]))*Long.valueOf(String.valueOf(b[i+1]))
                    *Long.valueOf(String.valueOf(b[i+2]))*Long.valueOf(String.valueOf(b[i+3]))
                    *Long.valueOf(String.valueOf(b[i+4]))*Long.valueOf(String.valueOf(b[i+5]))
                    *Long.valueOf(String.valueOf(b[i+6]))*Long.valueOf(String.valueOf(b[i+7]))
                    *Long.valueOf(String.valueOf(b[i+8]))*Long.valueOf(String.valueOf(b[i+9]))
                    *Long.valueOf(String.valueOf(b[i+10]))*Long.valueOf(String.valueOf(b[i+11]))
                    *Long.valueOf(String.valueOf(b[i+12])));
        }
        return list;
    }
    /*
    *  遍歷ArrayList數組
    *  比較輸出最大值
    */
    public static long num(List list)
    {
        long max=(Long)list.get(0);
        for(int i=0;i<list.size();i++)
        {
            if(max < (Long)list.get(i))
            {
                max = (Long)list.get(i);
            }
        }
        return max;
    }
}

Answer : 23514624000

時間效率： 5 毫秒。