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 = 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?

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

class Solution {
    public int combinationSum4(int[] nums, int target) {
        int[] res = new int[1];
        help(res, new ArrayList(), nums, target, 0);
        return res[0];
    }
    public void help(int[] res, List<Integer> cur, int[] nums, int target, int sum) {
        if(sum > target) return;
        if(sum == target) {
            res[0]++;
            return;
        }
        
        for(int i = 0; i < nums.length; i++) {
            sum += nums[i];
            cur.add(nums[i]);
            help(res, cur, nums, target, sum);
            sum -= nums[i];
            cur.remove(cur.size() - 1);
        }
    }
}

TLE了，屮

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

这个也是递归，也tle了。

仔细看一下，和coin change2十分相似，都是从中间态得到最终结果，那就是个dp，还是初始化dp【0】 = 1

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

看清楚需求，这个包括了重复的，所以coin在inner loop