39 Combination Sum
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.
- 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]
]
Solution:
class Solution {
public:
int sum(vector<int>& nums) {
int res = 0, n = nums.size();
if ( n == 0 ) return res;
for ( int i = 0; i <= n-1; i++ ) res += nums[i];
return res;
}
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
int n = candidates.size();
vector<vector<int>> res;
vector<vector<int>> tmp_res1;
vector<int> tmp;
tmp_res1.push_back(tmp);
for ( int i = 0; i <= n-1; i++ ) {
int nres = tmp_res1.size();
for ( int j = 0; j <= nres-1; j++ ) {
tmp = tmp_res1[j];
int s = sum(tmp_res1[j]);
int icount = (target-s)/candidates[i];
for ( int k = 1; k <= icount; k++) {
tmp.push_back(candidates[i]);
if ( s + candidates[i]*k == target ) res.push_back(tmp);
else tmp_res1.push_back(tmp);
}
}
}
return res;
}
};
Recursive Solution
class Solution {
public:
void help(vector<int>& candidates, int i0, vector<int> curr, int target, vector<vector<int>>& res) {
if ( target == 0 ) {res.push_back(curr); return;}
int n = candidates.size();
if ( i0 == n ) return;
while ( target >= 0 ) {
help(candidates, i0+1, curr, target, res);
curr.push_back(candidates[i0]);
target -= candidates[i0];
}
return;
}
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
vector<vector<int>> res;
vector<int> curr;
help(candidates, 0, curr, target, res);
return res;
}
};