Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
Note:
- All numbers (including target) will be positive integers.
- Elements in a combination (a1, a2, � , ak) must be in non-descending order. (ie, a1 ? a2 ? � ? ak).
- The solution set must not contain duplicate combinations.
For example, given candidate set 2,3,6,7
and target 7
,
A solution set is: [7]
[2, 2, 3]
class Solution {
public:
vector<vector<int> > res;
int n;
vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
vector<int> r;
res.clear();
n = candidates.size();
sort(candidates.begin(), candidates.end());
gen(r,0,0, candidates, target);
return res;
}
void gen(vector<int> &r,int sum,int l, const vector<int>& can, int tar) {
if (sum == tar) {
res.push_back(r);
return;
}
if (sum > tar) return ;
for (int i = l; i < n; i++) {
r.push_back(can[i]);
sum += can[i];
gen(r, sum,i, can, tar);
sum -= can[i];
r.pop_back();
}
}
};
@2013-10-05
class Solution {
public:
vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
vector<vector<int> > res;
if (candidates.size() == 0) return res;
vector<int> v;
sort(candidates.begin(), candidates.end());
gen(candidates, res, v, target, 0);
return res;
}
void gen(vector<int>& can, vector<vector<int> >& res, vector<int>& v, int left, int cur) {
if (left == 0) {
res.push_back(v);
return;
}
if (left < 0 || cur >= can.size()) return;
int next = cur + 1;
while (next < can.size() && can[next] == can[cur]) next++;
int t = left, cnt = 0;
gen(can, res, v, t, next);
while (t > 0) {
v.push_back(can[cur]);
t -= can[cur];
cnt ++;
gen(can, res, v, t, next);
}
while (cnt-- > 0) v.pop_back();
}
};
Combination Sum II
Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
Each number in C may only be used once in the combination.
Note:
- All numbers (including target) will be positive integers.
- Elements in a combination (a1, a2, � , ak) must be in non-descending order. (ie, a1 ? a2 ? � ? ak).
- The solution set must not contain duplicate combinations.
For example, given candidate set 10,1,2,7,6,1,5
and target 8
,
A solution set is: [1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]
class Solution {
public:
vector<vector<int> > res;
int n;
vector<vector<int> > combinationSum2(vector<int> &candidates, int target) {
vector<int> r;
res.clear();
n = candidates.size();
sort(candidates.begin(), candidates.end());
gen(r,0,0, candidates, target);
return res;
}
void gen(vector<int> &r,int sum,int l, const vector<int>& can, int tar) {
if (sum == tar) {
res.push_back(r);
return;
}
if (sum > tar) return ;
for (int i = l; i < n; i++) {
if (i > l && can[i] == can[i-1]) {continue;}
r.push_back(can[i]);
sum += can[i];
gen(r, sum, i+1, can, tar);
sum -= can[i];
r.pop_back();
}
}
};
@2013-10-05
class Solution {
public:
vector<vector<int> > combinationSum2(vector<int> &num, int target) {
vector<vector<int> > res;
if (num.size() == 0) return res;
vector<int> v;
sort(num.begin(), num.end());
gen(num, target, res, v, 0);
return res;
}
void gen(const vector<int> &a, int tar, vector<vector<int> >& res, vector<int>& v, int cur) {
if (tar == 0) {
res.push_back(v);
return;
}
if (tar < 0 || cur >= a.size()) return;
int next = cur + 1;
while (next < a.size() && a[next] == a[cur]) next++;
gen(a,tar, res, v, next);
if (a[cur] <= tar) {
v.push_back(a[cur]);
gen(a, tar - a[cur], res, v, cur + 1);
v.pop_back();
}
}
};