0216. Combination Sum III (M)

Combination Sum III (M)

题目

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]]

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题意

从1-9这9个数中选取k个数，使其之和正好为n，求出所有这样的组合。

思路

排列组合题，使用回溯法很容易求解。

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代码实现

Java

class Solution {
    public List<List<Integer>> combinationSum3(int k, int n) {
        List<List<Integer>> ans = new ArrayList<>();
        dfs(n, k, 0, 1, new ArrayList<>(), ans);
        return ans;
    }

    private void dfs(int n, int k, int sum, int start, List<Integer> temp, List<List<Integer>> ans) {
        if (temp.size() == k && sum == n) {
            ans.add(new ArrayList<>(temp));
            return;
        }
      
        if (temp.size() >= k || sum >= n) {
            return;
        }
      
        for (int i = start; i <= 9; i++) {
            temp.add(i);
            dfs(n, k, sum + i, i + 1, temp, ans);
            temp.remove(temp.size() - 1);
        }
    }
}

JavaScript

/**
 * @param {number} k
 * @param {number} n
 * @return {number[][]}
 */
var combinationSum3 = function (k, n) {
  let ans = []
  dfs(0, [], 1, ans, k, n)
  return ans
}

let dfs = function (sum, tmp, begin, ans, k, n) {
  if (tmp.length === k && sum === n) {
    return ans.push([...tmp])
  }

  if (tmp.length >= k || sum >= n) {
    return
  }

  for (let i = begin; i <= 9; i++) {
    tmp.push(i)
    dfs(sum + i, tmp, i + 1, ans, k, n)
    tmp.pop()
  }
}