leetcode：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.
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]

------------------------------------------------------------------------------------------------------------------------

首先给定的数组元素是升序排列的，结果集合为：res; 中间结果集合为：partRes; 中间结果集合中元素和sum

所以开始时：元素和sum=0, 从第一个元素开始遍历；

partRes.push_back(当前处理元素)； sum+=当前处理元素；

当 sum=target时，res.push_back(partRes)；

当sum> targe时，弹出partRes中之前加进去的元素；

处理下一个元素；

具体代码如下：

void findSumPath(vector<int> candidates, int index, int sum, vector<int>& partRes, vector<vector<int>>& res, int target)
{
if (sum > target)
return;
if (sum == target)
{
res.push_back(partRes);
return;
}
for (int i = index; i < candidates.size(); ++i)
{
partRes.push_back(candidates[i]);
findSumPath(candidates, i, sum + candidates[i], partRes, res, target);
partRes.pop_back();
}
}

vector<vector<int> > combinationSum(vector<int> &candidates, int target)
{
vector<int> partRes;
vector<vector<int>> res;
findSumPath(candidates, 0, 0, partRes, res, target);
return res;
}

法二：

灵感来自于“凑硬币”问题：

vector<vector<int>> combinationSum(vector<int>& candidates, int target)
{
int alen = candidates.size();
vector<vector<int>> res;
vector<int> partRes;
sort(candidates.begin(), candidates.end());
process(res, partRes, candidates, alen, 0, target);
return res;
}
void process(vector<vector<int>>& res, vector<int>& partRes, vector<int> arr, int alen, int index, int target)
{
if (index == alen)
{
if (target == 0)
res.push_back(partRes);
else
return;
}
else
{
for (int i = 0; arr[index] * i <= target; ++i)
{
for (int j = 0; j < i; j++)
{
partRes.push_back(arr[index]);
}

process(res, partRes, arr, alen, index + 1, target - arr[index] * i);
for (int j = 0; j < i; ++j)
{
partRes.pop_back();
}

}
}
}

法三：

class Solution {
private:
vector<vector<int> > ret;
vector<int> a;
public:
void solve(int dep, int maxDep, int target, vector<int> &cand)
{
if (target < 0)
return;

if (dep == maxDep)
{
if (target == 0)
{
vector<int> res;
for(int i = 0; i < maxDep; i++)
for(int j = 0; j < a[i]; j++)
res.push_back(cand[i]);
ret.push_back(res);
}
return;
}

for(int i = 0; i <= target / cand[dep]; i++)
{
a[dep] = i;
solve(dep + 1, maxDep, target - cand[dep] * i, cand);
}
}

vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
sort(candidates.begin(), candidates.end());

a.resize(candidates.size());
ret.clear();
if (candidates.size() == 0)
return ret;

solve(0, candidates.size(), target, candidates);

return ret;
}
};