39. Combination Sum - Medium-优快云博客

本文深入探讨了如何使用回溯法解决给定候选数集合和目标数，寻找所有可能的组合，使组合内的数之和等于目标数的问题。通过详细的代码示例，展示了算法的具体实现过程，包括剪枝优化策略，以及复杂度分析。

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

The same repeated number may be chosen from candidates unlimited number of times.

Note:

All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

backtracking

每次递归到某个元素的时候，tmp中加入当前元素，target减去当前元素，当target=0的时候，把tmp加入res中

剪枝：排序+递归过程中判断当前元素是否大于target (大于就break)

s = target / min(nums[i]), T = C(s, 1) + C(s, 2) + ... + C(s, s) = 2^s -> O(2^s)

time: O(2^s), space: O( target / min(nums[i]) )

class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<>();
        Arrays.sort(candidates);
        backtracking(candidates, target, 0, new ArrayList<>(), res);
        return res;
    }
    
    private void backtracking(int[] candidates, int target, int idx, List<Integer> tmp, List<List<Integer>> res) {
        if(target == 0) res.add(new ArrayList<>(tmp));
        
        for(int i = idx; i < candidates.length; i++) {
            if(candidates[i] > target) break;
            tmp.add(candidates[i]);
            backtracking(candidates, target - candidates[i], i, tmp, res);
            tmp.remove(tmp.size() - 1);
        }
    }
}