[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);  
        }
    }
}