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：

參考自：http://www.cnblogs.com/grandyang/p/5705750.html

這道題的真正解法應該是用DP來做，解題思想有點像之前爬梯子的那道題Climbing Stairs，我們需要一個一維數組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數組，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，參見代碼如下：

int combinationSum4(vector<int>& nums, int target)
     {
         sort(nums.begin(), nums.end());
         vector<int>dp(target+1,0);
         dp[0] = 1;
         for (int i = 1; i <= target;i++)
         {
             for (int a : nums)
             {
                 if (i<a)break
 
;//a排序後只能越來越大 i>=a才有意義 否則提前結束內部循環
                 dp[i] += dp[i - a];
             }
         }
         return dp[target];
     }

[leetcode-377-Combination Sum IV]