377. Combination Sum IV

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 = 4The 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? 

Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.

动态规划解题。第一种方法是自底向上，建立res[target+1]数组保存能到达n的方式，对于每一个res，遍历nums，如果i-nums[j]>=0，则res[i] += res[i-nums[j]]。代码如下：

public class Solution {    public int combinationSum4(int[] nums, int target) {        int[] res = new int[target + 1];        res[0] = 1;        for (int i = 1; i < res.length; i ++) {            for (int j = 0; j < nums.length; j ++) {                if (i - nums[j] >= 0) {                    res[i] += res[i - nums[j]];                }            }        }        return res[target];    }}

第二种方法是自上而下，每次的targrt是target-nums[i]，进行递归。代码如下：

private int[] dp;public int combinationSum4(int[] nums, int target) {    dp = new int[target + 1];    Arrays.fill(dp, -1);    dp[0] = 1;    return helper(nums, target);}private int helper(int[] nums, int target) {    if (dp[target] != -1) {        return dp[target];    }    int res = 0;    for (int i = 0; i < nums.length; i++) {        if (target >= nums[i]) {            res += helper(nums, target - nums[i]);        }    }    dp[target] = res;    return res;}