39. Combination Sum-蒲公英云

39. Combination Sum

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]
]

难度：medium

题目：给定一无重复元素的集合和一指定数，找出所有由数组内元素之和为该数的序列。每个元素都可以被无限次使用。
注意：所以元素都为正整数包括给定的整数。答案不允许有重复的组合。

思路：递归

Runtime: 9 ms, faster than 81.90% of Java online submissions for Combination Sum.
Memory Usage: 29.6 MB, less than 16.71% of Java online submissions for Combination Sum.

class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        Arrays.sort(candidates);
        List<List<Integer>> result = new ArrayList();
        combinationSum(candidates, 0, target, 0, new Stack<Integer>(), result);
        return result;
    }
    public void combinationSum(int[] cs, int idx, int target, int sum, Stack<Integer> stack, List<List<Integer>> result) {
        if (sum == target) {
            result.add(new ArrayList(stack));
            return;
        }
        for (int i = idx; i < cs.length; i++) {
            if (sum + cs[i] <= target) {
                stack.push(cs[i]);
                combinationSum(cs, i, target, sum + cs[i], stack, result);
                stack.pop();
            }
            if (sum + cs[i] >= target) break;
        }
    }
}