[leetcode]39. Combination Sum组合之和

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

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

Note:

• All numbers (including target) will be positive integers.
• The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

题意：

给定一个集合以及一个值target，找出所有加起来等于target的组合。（每个元素可以用无数次）

Solution1: Backtracking

code:

 /*
 Time: O(n!)   factorial,  n！=1×2×3×…×n
 Space: O(n)  coz n levels in stack for recrusion
 */

 class Solution {
     public List<List<Integer>> combinationSum(int[] nums, int target) {
         Arrays.sort(nums); // 呼应dfs的剪枝动作
         List<List<Integer>> result = new ArrayList<>();
         List<Integer> path = new ArrayList<>();
         dfs(nums, path, result, target, 0);
         return result;
     }

     private static void dfs(int[] nums, List<Integer> path,
                             List<List<Integer>> result, int remain, int start) {
         // base case
         if (remain == 0) {
             result.add(new ArrayList<Integer>(path));
             return;
         }

         for (int i = start; i < nums.length; i++) {
             if (remain < nums[i]) return; //基于 Arrays.sort(nums);
             path.add(nums[i]);
             dfs(nums, path, result, remain - nums[i], i);
             path.remove(path.size() - 1);
         }
     }
 }