216. Combination Sum III
Tags: ‘Array’, ‘Backtracking’
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 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]]
Solution
backtracking 模板 59% 6%
public List<List<Integer>> combinationSum3(int k, int n) {
int[] nums = {1,2,3,4,5,6,7,8,9};
List<List<Integer>> result = new ArrayList<>();
backtrack(result, new ArrayList<Integer>(), nums, k, n, 0);
return result;
}
private void backtrack(List<List<Integer>> result, List<Integer> temp, int[] nums, int countRemain, int remain, int start) {
// 结束条件
if (countRemain == 0 && remain == 0) {
result.add(new ArrayList<>(temp));
return;
} else if (remain < 0 || remain < 0) {
return;
}
for (int i = start; i < nums.length; i++) {
temp.add(nums[i]);
backtrack(result, temp, nums, countRemain - 1, remain - nums[i], i + 1);
temp.remove(temp.size() - 1);
}
}
方法2，可以不用nums 59% 6%
public List<List<Integer>> combinationSum3(int k, int n) {
List<List<Integer>> result = new ArrayList<>();
backtrack(result, new ArrayList<Integer>(), k, n, 1);
return result;
}
private void backtrack(List<List<Integer>> result, List<Integer> temp, int countRemain, int remain, int start) {
// 结束条件
if (countRemain == 0 && remain == 0) {
result.add(new ArrayList<>(temp));
return;
} else if (remain < 0 || remain < 0) {
return;
}
for (int i = start; i <= 9; i++) {
temp.add(i);
backtrack(result, temp, countRemain - 1, remain - i, i + 1);
temp.remove(temp.size() - 1);
}
}