【Rich Button】216. Combination Sum III
Find all combinations of k numbers that add up to n such that the following conditions are met:
only the digits 1 through 9 are used, each digit is used at most once, and a list of all possible valid combinations is returned. The list cannot contain the same combination twice, and the combinations may be returned in any order.
Example 1:
Input: k = 3, n = 7
Output: [[1,2,4]]
Explanation:
1 + 2 + 4 = 7
There is no other matching combination.
Example 2:
Input: k = 3, n = 9
Output: [[1,2,6], [1,3,5], [2,3,4]]
Explanation:
1 + 2 + 6 = 9
1 + 3 + 5 = 9
2 + 3 + 4 = 9
There is no other matching combination.
Example 3:
Input: k = 4, n = 1
Output: []
Explanation: No valid combination exists.
Using 4 different numbers in the range [1,9], the smallest sum we can get is 1+2+3+4 = 10, since 10 > 1, there is no valid combination.
提示:
2 <= k <= 9
1 <= n <= 60
answer
backtrace:
import java.util.*;
class Solution {
List<List<Integer>> result = new ArrayList<>();
LinkedList<Integer> path = new LinkedList<>();
public List<List<Integer>> combinationSum3(int k, int n) {
backTracking(n, k, 1, 0);
return result;
}
private void backTracking(int targetSum, int k, int startIndex, int sum) {
if (path.size() == k) {
if (sum == targetSum) {
result.add(new ArrayList<>(path));
}
return;
}
for (int i = startIndex; i <= 9 ; i++) {
path.add(i);
sum += i;
backTracking(targetSum, k, i + 1, sum);
//回溯
path.removeLast();
//回溯
sum -= i;
}
}
}
pruning
- If Sum is greater than TargetSum, there is no need to backtrack.
- In terms of quantity, if k numbers are required, there is no need to backtrack if there are not enough k numbers.
k - path.size()How many numbers are needed in the current layer?
import java.util.*;
class Solution {
List<List<Integer>> result = new ArrayList<>();
LinkedList<Integer> path = new LinkedList<>();
public List<List<Integer>> combinationSum3(int k, int n) {
backTracking(n, k, 1, 0);
return result;
}
private void backTracking(int targetSum, int k, int startIndex, int sum) {
// 减枝
if (sum > targetSum) {
return;
}
if (path.size() == k) {
if (sum == targetSum) {
result.add(new ArrayList<>(path));
}
return;
}
// 减枝 9 - (k - path.size()) + 1
for (int i = startIndex; i <= 9 - (k - path.size()) + 1; i++) {
path.add(i);
sum += i;
backTracking(targetSum, k, i + 1, sum);
//回溯
path.removeLast();
//回溯
sum -= i;
}
}
}