39. 组合总和


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,则两种组合是唯一的。 

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

Python:


class Solution:
    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
        res = []
        tem = []
        candidates.sort()

        def backStack(begin, tar):
            if tar == 0:
                res.append(tem[:])
            else:
                for i in range(begin, len(candidates)):
                    # 剪枝
                    if tar < candidates[i]:
                        break
                    tem.append(candidates[i])
                    backStack(i, tar - candidates[i])
                    tem.pop()

        backStack(0, target)
        return res

Java:

 List<List<Integer>> res = new ArrayList<>();
    List<Integer> tem = new ArrayList<>();
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        Arrays.sort(candidates);
        backStack(candidates, target, 0);
        return res;
    }

    private void backStack(int[] candidates, int temTarget, int i) {
        if (temTarget == 0) {
            res.add(new ArrayList<>(tem));
            return;
        } else if (temTarget < 0) {
            return;
        }
        for (int j = i; j < candidates.length; j++) {
            if (temTarget - candidates[j] < 0) {
                break;
            }
            tem.add(candidates[j]);
            backStack(candidates, temTarget - candidates[j], j);
            tem.remove(tem.size() - 1);
        }
    }
  • 作者:低调做个路人 (扫码联系作者)
  • 发表时间:2021-08-24 20:48:16
  • 版权声明:自由转载-非商用-非衍生-保持署名(创意共享3.0许可证)
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