题目描述：
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]
]
中文理解：
给定一个数组和指定值，然后数组中的值可以使用多次，给出数组中值加和为指定值的所有组合。
解题思路：
使用回溯的方法，具体详见代码。
代码（java）：
class Solution {
public List<List<Integer>> combinationSum(int[] candidates, int target) {
List<List<Integer>> list=new ArrayList();
Arrays.sort(candidates);
backtrace(list,new ArrayList<Integer>(),candidates,target,0);
return list;
}
public void backtrace(List<List<Integer>> list,List<Integer> tempList,int []nums,int remain,int start){
if(remain<0)return ;
else if(remain==0)list.add(new ArrayList<>(tempList));
else{
for(int i=start;i<nums.length;i++){
tempList.add(nums[i]);
backtrace(list,tempList,nums,remain-nums[i],i);
tempList.remove(tempList.size()-1);
}
}
}
}