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
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
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);
}
}
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