动态规划之子集和问题

原文地址：Dynamic Programming | Set 25 (Subset Sum Problem)

已知一个非负整数集，与sum的值，确定这个集合是否存在这样的子集，这个子集所有元素和等于sum。

例子: set[] = {3, 34, 4, 12, 5, 2}, sum = 9输出:  True  //There is a subset (4, 5) with sum 9.

设isSubSetSum(int set[], int n, int sum)就是这样的一个函数，它可以在set中找到一个子集，这个子集所有元素和是sum。n是set[]的元素个数。

isSubsetSum可以分为两个子问题：
…a)包括最后一个元素，用n = n-1, sum = sum – set[n-1]递归。
…b)不包括最后一个元素， 用n = n-1递归。

如果以上两个子问题任意一个返回true，那么整个问题返回true。

下面是isSubsetSum()问题的递归方程：

isSubsetSum(set, n, sum) = isSubsetSum(set, n-1, sum) ||                            isSubsetSum(set, n-1, sum-set[n-1])Base Cases:isSubsetSum(set, n, sum) = false, if sum > 0 and n == 0isSubsetSum(set, n, sum) = true, if sum == 0 

下面是根据上述的递归结构的一个简单的递归实现：

// A recursive solution for subset sum problemclass subset_sum{    // Returns true if there is a subset of set[] with sum        // equal to given sum    static boolean isSubsetSum(int set[], int n, int sum)    {       // Base Cases       if (sum == 0)         return true;       if (n == 0 && sum != 0)         return false;       // If last element is greater than sum, then ignore it       if (set[n-1] > sum)         return isSubsetSum(set, n-1, sum);       /* else, check if sum can be obtained by any of the following          (a) including the last element          (b) excluding the last element   */       return isSubsetSum(set, n-1, sum) ||                                    isSubsetSum(set, n-1, sum-set[n-1]);    }    /* Driver program to test above function */    public static void main (String args[])    {          int set[] = {3, 34, 4, 12, 5, 2};          int sum = 9;          int n = set.length;          if (isSubsetSum(set, n, sum) == true)             System.out.println("Found a subset with given sum");          else             System.out.println("No subset with given sum");    }}/* This code is contributed by Rajat Mishra */

输出：

Found a subset with given sum 

上述的解法在最坏情况下可能会尝试已知集合的所有子集。因此上述方法的时间复杂度是指数级的。这个问题实际上是NP-Complete(这个问题没有为止多项式的解法)。

这个问题可以在伪多项式时间内用动态规划的方法解决。我们可以建立一个boolean类型的二维数组subset[][]，然后用自底向上的方法填充。如果set[0..j-1]的一个子集的和等于i，那么subset[i][j]的值就为true，否则就是false。最后我们返回subset[sum][n]。

// A Dynamic Programming solution for subset sum problemclass subset_sum{    // Returns true if there is a subset of set[] with sun equal to given sum    static boolean isSubsetSum(int set[], int n, int sum)    {        // The value of subset[i][j] will be true if there             // is a subset of set[0..j-1] with sum equal to i        boolean subset[][] = new boolean[sum+1][n+1];        // If sum is 0, then answer is true        for (int i = 0; i <= n; i++)          subset[0][i] = true;        // If sum is not 0 and set is empty, then answer is false        for (int i = 1; i <= sum; i++)          subset[i][0] = false;         // Fill the subset table in botton up manner         for (int i = 1; i <= sum; i++)         {           for (int j = 1; j <= n; j++)           {             subset[i][j] = subset[i][j-1];             if (i >= set[j-1])               subset[i][j] = subset[i][j] ||                                           subset[i - set[j-1]][j-1];           }         }        /* // uncomment this code to print table         for (int i = 0; i <= sum; i++)         {           for (int j = 0; j <= n; j++)              printf ("%4d", subset[i][j]);           printf("\n");         } */         return subset[sum][n];    }    /* Driver program to test above function */    public static void main (String args[])    {          int set[] = {3, 34, 4, 12, 5, 2};          int sum = 9;          int n = set.length;          if (isSubsetSum(set, n, sum) == true)             System.out.println("Found a subset with given sum");          else             System.out.println("No subset with given sum");    }}/* This code is contributed by Rajat Mishra */

输出：

Found a subset with given sum

上述解法的时间复杂度是 O(sum*n)。