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.

一刷
題解：用傳統(tǒng)的backtrack會出現(xiàn)超時，由于題目暗示用dynamic programming, 容易聯(lián)想到轉(zhuǎn)移方程為：
f(target) = f(target - nums[0]) + f(target - nums[1]) + ... + f(target - nums.back())

public class Solution {
    public int combinationSum4(int[] nums, int target) {
        int[] dp = new int[target+1];
        dp[0] = 1;
        for(int i=1; i<=target; i++){
            for(int num:nums){
                if(i>=num) dp[i]+=dp[i-num];
            }
        }
        return dp[target];
    }
}

二刷
dynamic programming

class Solution {
    public int combinationSum4(int[] nums, int target) {
        int[] dp = new [target+1];
        dp[0] = 1;
        for(int i=1; i<=target; i++){
            for(int num : nums){
                if(i>=num) dp[i] += dp[i-num];
            }
        }
        return dp[target];
    }
}

共有4個combination sum:
Combination Sum
Combination Sum II
Combination Sum III
Combination Sum IV