377. Combination Sum IV

題目要求

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?

有一個不包含重複值的正整數數組nums，問從數組中選擇幾個數，其和爲target，這樣的數的組合有幾種？

思路一：自頂向下的dp

這題本質上需要注意一點，就是我如果需要組成target，那麼一定是由nums中的一個值和另一個值的排列組合結果構成的。比如com[4] = com[4-1] + com[4-2] + com[4-1]。通過這一點，我們構成一個遞歸表達式，但是因爲單純的遞歸表達式沒有計算中間結果，所以會造成大量重複的計算影響效率，所以這裏採用dp的思路額外的用數組來記錄已經計算過的com結果。比如com[3] = com[2] + com[1], com[2] = com[1]，如果沒有dp，則需要重複計算com[1]的結果。

    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];
    }