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.

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？

=============================

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。