216 Combination Sum III
Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.
Example 1:
Input:k= 3,n= 7
Output:
[[1,2,4]]
Example 2:
Input:k= 3,n= 9
Output:
[[1,2,6], [1,3,5], [2,3,4]]
这题跟前两题很相似,不过我们只要把k个元素装到tmp就ok了。所以递归结束条件是tmp.size() == k,另外我们还是找combination sum,所以还得判断一下target(n) 是否为0,如果是,才把tmp加到结果里。这里可以取的数从1到9 inclusive,所以loop的时候到=9。同样,会令target为负的,我们跳过。因为有序,所以可以break跳过后面全部。因为每个数字只能取一次,所以进入递归时传入i+1.
public List<List<Integer>> combinationSum3(int k, int n) {
List<List<Integer>> res = new ArrayList<>();
if (k == 0 || n < 0) {
return res;
}
ArrayList<Integer> tmp = new ArrayList<>();
dfsHelper(tmp, res, k, n, 1);// 从1到9 inclusive,所以从1开始取
return res;
}
private void dfsHelper(ArrayList<Integer> tmp, List<List<Integer>> res,
int k, int n, int start) {
if (tmp.size() == k) {// 要取k个数
if (n == 0) {// target要减到0
res.add(new ArrayList<>(tmp));
}
return;
}
for (int i = start; i <= 9; i++) {
if (i > n) {// 跳过会把target减为负数的数
break;
}
tmp.add(i);
// 每个数只能取1次,下次递归从i + 1开始
dfsHelper(tmp, res, k, n - i, i + 1);
tmp.remove(tmp.size() - 1);
}
}
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