组合总和

Algorithm

39. Combination Sum

Description

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.

It is guaranteed 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: []

Example 4:

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

Example 5:

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

Constraints:

• 1 <= candidates.length <= 30
• 1 <= candidates[i] <= 200
• All elements of candidates are distinct.
• 1 <= target <= 500

Solution

class Solution {
    public List> combinationSum(int[] candidates, int target) {
        List> list = new ArrayList<>();
        Arrays.sort(candidates);
        backtrack(list, new ArrayList<>(), candidates, target, 0);
        return list;
    }

    public void backtrack(List> list, List tempList, int [] nums, int remain, int start){
        if(remain < 0){
            return;
        }else if(remain==0){
            list.add(new ArrayList<>(tempList));
        }else{
            for(int i = start; i < nums.length; i++){
                tempList.add(nums[i]);
                // 从i开始因为可以有重复元素
                backtrack(list, tempList, nums, remain - nums[i], i);
                // 能够跳出循环说明当前remain<0, 需要去掉最后一个扣减的元素
                tempList.remove(tempList.size() - 1);
            }
        }
    }
}