Combination Sum III
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2022-05-29 18:22:37
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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]]
找出所有可能的组合,使组合中元素的个数等于k, 元素的和等于n。我们采用回溯法,当元素的个数为k并且和为9的时候我们记录这个结果,然后回溯,继续寻找其他可能的结果。代码如下:
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]]
找出所有可能的组合,使组合中元素的个数等于k, 元素的和等于n。我们采用回溯法,当元素的个数为k并且和为9的时候我们记录这个结果,然后回溯,继续寻找其他可能的结果。代码如下:
public class Solution { public List<List<Integer>> combinationSum3(int k, int n) { LinkedList<Integer> list = new LinkedList<Integer>(); List<List<Integer>> result = new ArrayList<List<Integer>>(); getCombinations(1, k, n, list, result); return result; } public void getCombinations(int start, int k, int n, LinkedList<Integer> list, List<List<Integer>> llist) { if(list.size() == k && n == 0) { llist.add(new LinkedList<Integer>(list)); } for(int i = start; i <= 9; i++) { if(list.size() < k) { list.add(i); getCombinations(i + 1, k, n - i, list, llist); list.removeLast(); } } } }
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