另一种看待这个问题的方法是找到子集和的数量。

// Create a map with key as the sum of a subset and value as the number of ways this sum was found.
Map<Integer, Integer> m = new HashMap<>(); 
// initialising as 0 can be found by selecting no numbers 
primary.put(0,0);
for (Integer i : numbers) {
  List<Integer> newSums = new ArrayList<>();
  for (Integer possibleSum : m.keySet()) {
        if (i+possibleSum > targetSum)
          continue;
        newSums.add(i+possibleSum);
  }
  for (Integer newPossibleSum : newSums) {
        m.put(newPossibleSum, m.getOrDefault(newPossibleSum, 0)+1);
  } 
}
return m.getOrDefault(targetSum,0);

我发现了我的错误，只是张贴了解决方案，以防有人面临同样的问题。

public int findNoOfSubset(int[] nums, int n, int sum) {
    if(sum == 0 && n<0){
        return 1;
    }

    if(n < 0 || sum < 0) {
        return 0;
    }

    if(dp[n][sum] != -1) {
        return dp[n][sum];
    }

    if(sum >= nums[n]) {
        return dp[n][sum] = findNoOfSubset(nums, n-1, sum-nums[n]) + findNoOfSubset(nums, n-1, sum);
    }

    return dp[n][sum] = findNoOfSubset(nums, n-1, sum);
}