377. Combination Sum IV

题目要求

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?

有一个不包含重复值的正整数数组nums，问从数组中选择几个数，其和为target，这样的数的组合有几种？

思路一：自顶向下的dp

这题本质上需要注意一点，就是我如果需要组成target，那么一定是由nums中的一个值和另一个值的排列组合结果构成的。比如com[4] = com[4-1] + com[4-2] + com[4-1]。通过这一点，我们构成一个递归表达式，但是因为单纯的递归表达式没有计算中间结果，所以会造成大量重复的计算影响效率，所以这里采用dp的思路额外的用数组来记录已经计算过的com结果。比如com[3] = com[2] + com[1], com[2] = com[1]，如果没有dp，则需要重复计算com[1]的结果。

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

思路二：自底向上dp

    public int combinationSum4(int[] nums, int target) {
        Arrays.sort(nums);
        int[] combinationCount = new int[target+1];
        
        combinationCount[0] = 1;
        
        for(int i = 1 ; i<=target ; i++) {
            for(int j = 0 ; j<nums.length && nums[j] <= i ; j++) {
                combinationCount[i] += combinationCount[i - nums[j]];
            }
        }
        return combinationCount[target];
    }

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