代码来自：https://www.geeksforgeeks.org/subset-sum-problem-dp-25/
    应重点记忆上面链接中第二种动态规划的方法，因为第一种方法是指数复杂度。这里只记录动态规划的算法。

    1. /*
    2. We can solve the problem in Pseudo-polynomial time using Dynamic programming. 
    3. We create a boolean 2D table subset[][] and fill it in bottom up manner. 
    4. The value of subset[i][j] will be true if there is a subset of set[0..j-1] with sum equal to i.,
    5. otherwise false. Finally, we return subset[sum][n]
    7. */
    10. // Returns true if there is a subset of  
    11.     // set[] with sun equal to given sum 
    12.     static boolean isSubsetSum(int set[],  
    13.                              int n, int sum) 
    14.     { 
    15.         // The value of subset[i][j] will be 
    16.         // true if there is a subset of  
    17.         // set[0..j-1] with sum equal to i 
    18.         boolean subset[][] =  
    19.                      new boolean[sum+1][n+1]; 
    21.         // If sum is 0, then answer is true 
    22.         for (int i = 0; i <= n; i++) 
    23.             subset[0][i] = true; 
    25.         // If sum is not 0 and set is empty, 
    26.         // then answer is false 
    27.         for (int i = 1; i <= sum; i++) 
    28.             subset[i][0] = false; 
    30.         // Fill the subset table in botton 
    31.         // up manner 
    32.         for (int i = 1; i <= sum; i++) 
    33.         { 
    34.             for (int j = 1; j <= n; j++) 
    35.             { 
    36.                 subset[i][j] = subset[i][j-1]; 
    37.                 if (i >= set[j-1]) 
    38.                 subset[i][j] = subset[i][j] ||  
    39.                      subset[i - set[j-1]][j-1]; 
    40.             } 
    41.         } 
    43.         /* // uncomment this code to print table 
    44.         for (int i = 0; i <= sum; i++) 
    45.         { 
    46.         for (int j = 0; j <= n; j++) 
    47.             System.out.println (subset[i][j]); 
    48.         } */
    50.         return subset[sum][n]; 
    51.     }