dfs/combination-sum-iii
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2023-12-01
Combination Sum III
描述
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.
Ensure that numbers within the set are sorted in ascending order.
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]]
分析
这是一个多阶段问题，目标是求所有解，显然用深搜+剪枝，即回溯法。
代码
// Combination Sum III
// Time Complexity: O(9*8*...*(10-k)), Space Complexity: O(k)
public class Solution {
public List<List<Integer>> combinationSum3(int k, int n) {
final List<List<Integer>> result = new ArrayList<>();
final List<Integer> path = new ArrayList<>();
dfs(k, n, path, result);
return result;
}
private static void dfs(int step, int gap, List<Integer> path,
List<List<Integer>> result) {
if (step == 0) {
if (gap == 0) {
result.add(new ArrayList<>(path));
}
return;
}
if (gap < 1) return;
final int start = path.isEmpty() ? 1 : path.get(path.size() - 1)+1;
for (int i = start; i < 10; ++i) {
path.add(i);
dfs(step - 1, gap - i, path, result);
path.remove(path.size() - 1);
}
}
}