- 回溯框架
- base case:总和等于时添加res;大于时return
- 使用 start 递归调用时把 i 自身传入即可实现重复取,又可以避免结果中相同的结果不同顺序出现的可能
class Solution {
List<List<Integer>> res = new LinkedList<>();
LinkedList<Integer> track = new LinkedList<>();
int sum = 0;
public List<List<Integer>> combinationSum(int[] candidates, int target) {
if(candidates.length == 0){
return res;
}
backtrack(candidates,target,0);
return res;
}
public void backtrack(int[] candidates,int target,int start){
if(target == sum){
res.add(new LinkedList(track));
return;
}
if(sum >target){
return;
}
for(int i = start;i<candidates.length;i++){
track.add(candidates[i]);
sum += candidates[i];
backtrack(candidates,target,i);
sum -= candidates[i];
track.removeLast();
}
}
}