Combination Sum

描述

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Note:

  • All numbers (including target) will be positive integers.
  • Elements in a combination (a1,a2,…,aka_1, a_2, …, a_ka​1​​,a​2​​,…,a​k​​) must be in non-descending order. (ie, a1≤a2≤…≤aka_1 \leq a_2 \leq … \leq a_ka​1​​≤a​2​​≤…≤a​k​​).
  • The solution set must not contain duplicate combinations.
    For example, given candidate set 2,3,6,7 and target 7, A solution set is: 
  1. [7] 
  2. [2, 2, 3]

分析

无

代码

  1. // Combination Sum
  2. // 时间复杂度O(n!)，空间复杂度O(n)
  3. public class Solution {
  4.     public List<List<Integer>> combinationSum(int[] nums, int target) {
  5.         Arrays.sort(nums);
  6.         List<List<Integer>> result = new ArrayList<>(); // 最终结果
  7.         List<Integer> path = new ArrayList<>(); // 中间结果
  8.         dfs(nums, path, result, target, 0);
  9.         return result;
  10.     }
  12.     private static void dfs(int[] nums, List<Integer> path, 
  13.                             List<List<Integer>> result, int gap, int start) {
  14.         if (gap == 0) {  // 找到一个合法解
  15.             result.add(new ArrayList<Integer>(path));
  16.             return;
  17.         }
  18.         for (int i = start; i < nums.length; i++) { // 扩展状态
  19.             if (gap < nums[i]) return; // 剪枝
  21.             path.add(nums[i]); // 执行扩展动作
  22.             dfs(nums, path, result, gap - nums[i], i);
  23.             path.remove(path.size() - 1);  // 撤销动作
  24.         }
  25.     }
  26. }

相关题目

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