Given an integer array with all positive numbers and no duplicates, find the number of possible combinations that add up to a positive integer target.
Example:
nums = [1, 2, 3]
target = 4
The possible combination ways are:
(1, 1, 1, 1)
(1, 1, 2)
(1, 2, 1)
(1, 3)
(2, 1, 1)
(2, 2)
(3, 1)
Note that different sequences are counted as different combinations.
Therefore the output is 7.
Follow up:
What if negative numbers are allowed in the given array?
How does it change the problem?
What limitation we need to add to the question to allow negative numbers?
思路：递归求解。
为了避免相同的解重复计数，要将原数组中的重复数字剔除，这样子所有的情况都只会枚举一遍。
同时，为了提速，在递归过程中，可以用一个map记录子问题的结果，这样就可以节省时间。
补充：如果数组中有负数，则应该添加的额外条件是最多可以有几个数相加。
1 class Solution {
2 public:
3 int help(vector<int>& nums, int target, unordered_map<int, int>& solutionCount) {
4 int count = 0;
5 for (int i = 0; i < nums.size() && nums[i] <= target; i++) {
6 if (nums[i] < target) {
7 int balance = target - nums[i];
8 if (solutionCount.count(balance))
9 count += solutionCount[balance];
10 else {
11 int subCount = help(nums, balance, solutionCount);
12 solutionCount.insert(make_pair(balance, subCount));
13 count += subCount;
14 }
15 }
16 else count++;
17 }
18 return count;
19 }
20 int combinationSum4(vector<int>& nums, int target) {
21 if (nums.size() == 0) return 0;
22 sort(nums.begin(), nums.end(), less<int>());
23 vector<int> distinctNum(1, nums[0]);
24 unordered_map<int, int> solutionCount;
25 for (int i = 1; i < nums.size(); i++)
26 if (nums[i] != nums[i-1]) distinctNum.push_back(nums[i]);
27 return help(distinctNum, target, solutionCount);
28 }
29 };