題目連結：

題目：

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：

dp，dp[i]表示當target為i 時，有多少種組合。 

狀態轉移方程：dp[i]=Σdp[i-nums[k]]  0<=k<=nums.length 

當然，需要考慮當i-nums[k]為0時，表示陣列中有target，則此時dp[i]為1,

時間複雜度O(n^2)

演算法：

	public int combinationSum4(int[] nums, int target) {
		int dp[] = new int[target + 1];
		for (int i = 0; i <= target; i++) {
			for (int k = 0; k < nums.length; k++) {
				if (i - nums[k] > 0) {
					dp[i] += dp[i - nums[k]];
				} else if (i - nums[k] == 0) {
					dp[i] += 1;
				}

			}
		}
		return dp[target];
	}