题目
解题思路
- 对数组进行排序;
- 递推寻找合适的元素;
- 当目标结果为0时添加到返回结果中;
- 当索引越界或目标值小于0时返回停止当前递推;
- 因为允许重复获取同一元素,所以分两种,一种下一元素还是当前元素,另一种为索引后推。
代码展示
class Solution {
List<List<Integer>> ans = new ArrayList<>();
int[] candidates;
int len;
public List<List<Integer>> combinationSum(int[] candidates, int target) {
Arrays.sort(candidates);
this.candidates = candidates;
len = candidates.length;
dfs(0,target, new ArrayList<>());
return ans;
}
private void dfs(int index, int target, List<Integer> list){
if(target == 0){
ans.add(new ArrayList<>(list));
return;
}
if(index >= len || target < 0){
return;
}
if(candidates[index] <= target){
list.add(candidates[index]);
dfs(index, target - candidates[index], list);
list.remove(list.size() - 1);
}
dfs(index + 1, target, list);
}
}