Given a set of candidate numbers (C) (without duplicates) 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.
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]
]
class Solution {
public List<List<Integer>> combinationSum(int[] candidates, int target) {
Arrays.sort(candidates);//排序
List<List<Integer>> result = new ArrayList<>();
if(candidates.length == 0||target < candidates[0])
return result;
List<Integer> com = new ArrayList<>();
combine(candidates,com,target,result,0);
return result;
}
public void combine(int[] candidates,List<Integer> com,int target,List<List<Integer>> result,int begin){
if(target<candidates[begin]) return;//全是正数的情况下，说明已经不存在可行解了
for(int i = begin;i < candidates.length;i++){//i从begin开始，是为了避免找到重复解
if(target==candidates[i]){//找到可行解
com.add(candidates[i]);
result.add(new ArrayList<>(com));
com.remove(com.size()-1);//每次添加一个元素后，无论是不是可行，都要把这个元素再去掉，因为在现有的com里再加入其他元素也可能是可行解
return;
}
com.add(candidates[i]);//把元素加入com，在此情况下，调用combine计算是否可以继续添加元素得到可行解
combine(candidates,com,target-candidates[i],result,i);
com.remove(com.size()-1);
}
}
}
