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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(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)。
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