Combination Sum III

描述

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]]

分析

这是一个多阶段问题，目标是求所有解，显然用深搜+剪枝，即回溯法。

代码

  1. // Combination Sum III
  2. // Time Complexity: O(9*8*...*(10-k)), Space Complexity: O(k)
  3. public class Solution {
  4.     public List> combinationSum3(int k, int n) {
  5.         final List> result = new ArrayList<>();
  6.         final List path = new ArrayList<>();
  7.         dfs(k, n, path, result);
  8.         return result;
  9.     }
  11.     private static void dfs(int step, int gap, List path,
  12.                             List> result) {
  13.         if (step == 0) {
  14.             if (gap == 0) {
  15.                 result.add(new ArrayList<>(path));
  16.             }
  17.             return;
  18.         }
  20.         if (gap < 1) return;
  22.         final int start = path.isEmpty() ? 1 : path.get(path.size() - 1)+1;
  23.         for (int i = start; i < 10; ++i) {
  24.             path.add(i);
  25.             dfs(step - 1, gap - i, path, result);
  26.             path.remove(path.size() - 1);
  27.         }
  28.     }
  29. }

相关题目

原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/java/dfs/combination-sum-iii.html