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?

LeetCode：連結

這道題是說給的陣列中全是整數且沒有重複，但是每個數字可以重複使用，讓給出組合的總數，就不用給出所有的排列。沿用回溯的方法當然可以啦，然後求個result的長度就行了。But他是會超時的，這就是我說的給出有多少種的時候就用動態規劃

我們需要一個一維陣列dp，其中dp[i]表示目標數為i的解的個數，然後我們從1遍歷到target，對於每一個數i，遍歷nums陣列，如果i>=x, dp[i] += dp[i - x]。這個也很好理解，比如說對於[1,2,3] 4，這個例子，當我們在計算dp[3]的時候，3可以拆分為1+x，而x即為dp[2]，3也可以拆分為2+x，此時x為dp[1]，3同樣可以拆為3+x，此時x為dp[0]，我們把所有的情況加起來就是組成3的所有情況了。

class Solution(object):
    def combinationSum4(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        dp = [0] * (target + 1)
        dp[0] = 1
        for i in range(target + 1):
            for num in nums:
                if i >= num:
                    dp[i] += dp[i-num]
        return dp[target]
a = Solution()
print(a.combinationSum4([1,2,3], 4))