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：

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

这道题的真正解法应该是用DP来做，解题思想有点像之前爬梯子的那道题Climbing Stairs，我们需要一个一维数组dp，其中dp[i]表示目标数为i的解的个数，然后我们从1遍历到target，对于每一个数i，遍历nums数组，如果i>=x, dp[i] += dp[i - x]。这个也很好理解，比如说对于[1,2,3] 4，这个例子，当我们在计算dp[3]的时候，3可以拆分为1+x，而x即为dp[2]，3也可以拆分为2+x，此时x为dp[1]，3同样可以拆为3+x，此时x为dp[0]，我们把所有的情况加起来就是组成3的所有情况了，参见代码如下：

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排序后只能越来越大 i>=a才有意义 否则提前结束内部循环
                 dp[i] += dp[i - a];
             }
         }
         return dp[target];
     }

[leetcode-377-Combination Sum IV]