题目介绍

解法

是216组合总和III链接的扩展

class Solution {
    List<List<Integer>> res = new ArrayList<>();
    List<Integer> path = new ArrayList<>();
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        dfs(candidates, target, 0, 0);
        return res;
    }
    private void dfs(int[] candidates, int target, int startIndex, int sum) {
        if (sum >= target) {
            if (sum == target) {
                res.add(new ArrayList<>(path));
            }
            return;
        }
        for (int i = startIndex; i < candidates.length; i++) {
            path.add(candidates[i]);
            sum += candidates[i];
		    // 关键点:不用i+1了，表示可以重复读取当前的数
            dfs(candidates, target, i, sum);
            path.remove(path.size() - 1);
            sum -= candidates[i];
        }
    }
}