leetcode 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.
Ensure that numbers within the set are sorted in ascending order.

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]]

尼玛，不知道题目为什么要限制只能用1-9，表示不理解。开始按找全部组合做的，按题目要求过滤一下就行。

class Solution {
void do_once(int k,int n, map<set<int>, int>&candi)
{
map<set<int>, int>newcandi;
for (map<set<int>, int>::iterator it = candi.begin(); it != candi.end(); it++)
{
if (k == 1)
{
if (n - it->second > 0 && it->first.find(n - it->second) == it->first.end())
{
set<int>nums = it->first;
nums.insert(n - it->second);
newcandi[nums] = n;
}
}
else
for (int i = 1; i <= n - it->second-k+1; i++)
{
if (it->second+i<=n&&it->first.find(i) == it->first.end())
{
set<int>nums = it->first;
nums.insert(i);
newcandi[nums] = it->second +i;
}
}
}
candi = newcandi;
}
public:
vector<vector<int>> combinationSum3(int k, int n) {
vector<vector<int>>re;
map<set<int>,int>candi;
set<int>aa;
candi[aa] = 0;
while (k>0)
{
do_once(k,n, candi);
k--;
}
for (map<set<int>, int>::iterator it = candi.begin(); it != candi.end(); it++)
if (it->second == n)
re.push_back(vector<int>(it->first.begin(), it->first.end()));
return re;
}
};

这个太慢，不能满意

class Solution {
int sum(vector<int>nums, int startpos, int endpos)
{
int re = 0;
for (int j = startpos; j <= endpos; j++)
re += nums[j];
return re;
}
public:
vector<vector<int>> combinationSum3(int k, int n) {
vector<vector<int>>re;
if (k == 0)
return re;
if (k == 1)
{
vector<int>aa;
aa.push_back(n);
if (n <= 9)
re.push_back(aa);
return re;
}
vector<int>nums(k);
for (int i = 0; i < k - 1; i++)
nums[i] = i + 1;
int index = k - 1;
while (index >= 1)
{
int jj = index;
while (jj != k - 1)
{
nums[jj] = nums[jj - 1] + 1;
jj++;
}
nums[k - 1] = n - sum(nums, 0, k - 2);
while (nums[k - 1] > nums[k - 2])
{
bool flag = true;
for (int i = k - 1; i >= 0; i--)
if (nums[i] > 9)
{
flag = false;
break;
}
if (flag)
re.push_back(nums);
nums[k - 1]--;
nums[k - 2]++;
}
int pp = 3;
while (k - pp >= 0)
{
nums[k - pp]++;
if (nums[k - pp] * pp + pp*(pp - 1) / 2 <= n - sum(nums, 0, k - pp - 1))
{
index = k - pp + 1;
break;
}
else
pp++;
}
if (k - pp < 0)
break;
}
return re;
}
};

accepted