216 Combination Sum III 组合总和 III
Description:
Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.
Note:
All numbers will be positive integers.
The solution set must not contain duplicate combinations.
Example:
Example 1:
Input: k = 3, n = 7
Output: [[1,2,4]]
Example 2:
Input: k = 3, n = 9
Output: [[1,2,6], [1,3,5], [2,3,4]]
题目描述:
找出所有相加之和为 n 的 k 个数的组合。组合中只允许含有 1 - 9 的正整数，并且每种组合中不存在重复的数字。
说明：
所有数字都是正整数。
解集不能包含重复的组合。
示例 :
示例 1:
输入: k = 3, n = 7
输出: [[1,2,4]]
示例 2:
输入: k = 3, n = 9
输出: [[1,2,6], [1,3,5], [2,3,4]]
思路:
回溯法
终点为长度等于 n且 target == 0
时间复杂度O(n!), 空间复杂度O(n)
代码:
C++:
class Solution
{
public:
vector> combinationSum3(int k, int n)
{
vector> result;
vector track;
trackback(result, track, k, 1, n);
return result;
}
private:
void trackback(vector>& result, vector& track, int k, int start, int target)
{
if (target < 0) return;
if (track.size() == k and target == 0)
{
result.push_back(track);
return;
}
for (int i = start; i < 10; i++)
{
track.push_back(i);
trackback(result, track, k, i + 1, target - i);
track.pop_back();
}
}
};
Java:
class Solution {
public List> combinationSum3(int k, int n) {
List> result = new ArrayList<>();
trackback(result, new ArrayList(), k, 1, n);
return result;
}
private void trackback(List> result, List temp, int k, int start, int target) {
if (target < 0) return;
if (temp.size() == k && target == 0) {
result.add(new ArrayList<>(temp));
return;
}
for (int i = start; i < 10; i++) {
temp.add(i);
trackback(result, temp, k, i + 1, target - i);
temp.remove(temp.size() - 1);
}
}
}
Python:
class Solution:
def combinationSum3(self, k: int, n: int, r=range(1, 10)) -> List[List[int]]:
return [list(i) for i in itertools.combinations(range(1, 10), k) if sum(i) == n]