Find all valid combinations of k numbers that sum up to n such that the following conditions are true:
Each number is used at most once
.Return a list of all possible valid combinations. The list must not contain the same combination twice, and the combinations may be returned in any order.
Example 1:
Input: k = 3, n = 7Output: [[1,2,4]]Explanation:1 + 2 + 4 = 7There are no other valid combinations.
Example 2:
Input: k = 3, n = 9Output: [[1,2,6],[1,3,5],[2,3,4]]Explanation:1 + 2 + 6 = 91 + 3 + 5 = 92 + 3 + 4 = 9There are no other valid combinations.
Example 3:
Input: k = 4, n = 1Output: []Explanation: There are no valid combinations. [1,2,1] is not valid because 1 is used twice.
Example 4:
Input: k = 3, n = 2Output: []Explanation: There are no valid combinations.
Example 5:
Input: k = 9, n = 45Output: [[1,2,3,4,5,6,7,8,9]]Explanation:1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45There are no other valid combinations.
Constraints:
从题目 Each number is used at most once
得知, 该题属于组合问题。
/*** @param {number} k* @param {number} n* @return {number[][]}*/var combinationSum3 = function(k, n) {const result = []recursive(k, n, 1, [], result)return result};var recursive = function(k, n, start, temp, result) {if (n < 0 || temp.length > k) returnif (n === 0 && temp.length === k) {result.push([...temp])return}for (let i = start; i <= 9; i++) {temp.push(i)n = n - irecursive(k, n, i + 1, temp, result)n = n + itemp.pop()}}
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