Combination Sum - Leetcode

public class Solution {
   public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> result = new ArrayList<>();
        List<Integer> path  = new ArrayList<>();
        Arrays.sort(candidates);
        DFS(result, path, candidates, target, 0);
        return result;
    }
    private void DFS(List<List<Integer>> result, List<Integer> path, int[] pool, int gap, int start){
        if(gap == 0){
            result.add(new ArrayList<>(path));
            return;
        }
        for(int i=start; i<pool.length; i++){
            if(gap < pool[i])
               return;
            path.add(pool[i]);
            DFS(result, path, pool, gap-pool[i], i);
            path.remove(path.size()-1);
        }
    }
}

思路： 大概的思路就是从第一个元素开始遍历，与target进行比较，如果gap为0，则说明找到解；否则（gap不为0）再从第一个元素开始验证求解。

分析：每一个元素都做的深度挖掘和验证， DFS

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, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
• The solution set must not contain duplicate combinations.

For example, given candidate set 2,3,6,7 and target 7, 
A solution set is: 
[7] 
[2, 2, 3]