39. Combination Sum

Problem

Given an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the combinations in any order.

The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different.

The test cases are generated such that the number of unique combinations that sum up to target is less than 150 combinations for the given input.

Example 1:

Input: candidates = [2,3,6,7], target = 7
Output: [[2,2,3],[7]]
Explanation:
2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times.
7 is a candidate, and 7 = 7.
These are the only two combinations.

Example 2:

Input: candidates = [2,3,5], target = 8
Output: [[2,2,2,2],[2,3,3],[3,5]]

Example 3:

Input: candidates = [2], target = 1
Output: []

Constraints:

• 1 <= candidates.length <= 30
• 2 <= candidates[i] <= 40
• All elements of candidates are distinct.
• 1 <= target <= 40

Solution

此題要我們列出所有可能的組合，所以我們可以用 backtracking 來找。但是，我們必須避免找到重複的組合。當兩個組合裡的數字和數量都依一樣，但順序不一樣時，他們依然是相同的組合。為了避免這樣的情況發生，當我們每次在找下一個數字時，不可以從 candidates[0] 開始找，我們應該從當下的數字或是下一個數字開始找。然而，問題中有提到，數字是可以重複地使用。所以，我們應該從當下的數字開始找。

Backtracking

• Time:
• Space:

class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        if (candidates.length == 0) return new ArrayList<>();
        Set<List<Integer>> results = new HashSet<>();
        combinationSum(candidates, target, 0, new ArrayList<>(), results);
        return new ArrayList<>(results);
    }

    private void combinationSum(int[] candidates, int target, int index, List<Integer> prefix, Set<List<Integer>> results) {
        if (target < 0) return;

        if (target == 0) {
            results.add(new ArrayList<>(prefix));
            return;
        }

        for (int i = index; i < candidates.length; i++) {
            prefix.add(candidates[i]);
            combinationSum(candidates, target - candidates[i], i, prefix, results);
            prefix.remove(prefix.size() - 1);
        }
    }
}

參考

• 39. Combination Sum, LeetCode.
• 39. Combination Sum, LeetCode Solutions.