给定一个无重复元素的正整数数组 candidates
和一个正整数 target
,找出 candidates
中所有可以使数字和为目标数 target
的唯一组合。
candidates
中的数字可以无限制重复被选取。如果至少一个所选数字数量不同,则两种组合是不同的。
对于给定的输入,保证和为 target
的唯一组合数少于 150
个。
示例 1:
输入: candidates = [2,3,6,7], target = 7
输出: [[7],[2,2,3]]
示例 2:
输入: candidates = [2,3,5], target = 8
输出: [[2,2,2,2],[2,3,3],[3,5]]
示例 3:
输入: candidates = [2], target = 1
输出: []
示例 4:
输入: candidates = [1], target = 1
输出: [[1]]
示例 5:
输入: candidates = [1], target = 2
输出: [[1,1]]
提示:
1 <= candidates.length <= 30
1 <= candidates[i] <= 200
candidate
中的每个元素都是独一无二的。1 <= target <= 500
//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
public List<List<Integer>> combinationSum(int[] candidates, int target) {
Arrays.sort(candidates);
Set<List<Integer>> result = new HashSet<>();
List<Integer> curRes = new ArrayList<>();
combine(result, curRes, 0, 0, candidates, target);
return new ArrayList<>(result);
}
void combine(Set<List<Integer>> result, List<Integer> curRes, int index, int curSum, int[] candidates, int target) {
if (curSum == target) {
List<Integer> one = new ArrayList<>(curRes);
result.add(one);
return;
} else if (curSum > target || index >= candidates.length) {
return;
}
curRes.add(candidates[index]);
combine(result, curRes, index, curSum + candidates[index], candidates, target);
combine(result, curRes, index + 1, curSum + candidates[index], candidates, target);
curRes.remove(curRes.size() - 1);
combine(result, curRes, index + 1, curSum, candidates, target);
}
}
//leetcode submit region end(Prohibit modification and deletion)