39. 组合总和
题目描述
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 (a 1, a 2, … , a k) must be in non-descending order(非降序，即升序). (ie, a1 ≤ a2 ≤ … ≤ a k).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]
思路
前缀+一个元素，一层一层的产生，同一前缀加一个不同的元素，不同前缀之间由于前缀不同自然加同一个元素产生的一定不同
代码实现
class Solution {
public:
vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
vector<vector<int>> ret;
sort(candidates.begin(),candidates.end());
//显然我们不能使用原候选集合作为临时结果集合，所以这里需要我们自己定义
vector<int> temp;
int curSum = 0;
combinationSumCore(candidates,0,ret,temp,curSum,target);
return ret;
}
void combinationSumCore(vector<int> &candidates,int start,vector<vector<int>> &ret,vector<int> &temp,int &curSum,int target)
{
//这里的程序设计保证在进入递归函数时curSum一定没有达到目标
for(int i = start;i < candidates.size();i++)
{
curSum += candidates[i];
if(curSum > target)
{
curSum-=candidates[i];
return;//剪枝
}
else if(curSum == target)
{
temp.push_back(candidates[i]);
ret.push_back(temp);
temp.pop_back();
curSum-=candidates[i];
return;//剪枝
}
else
{
temp.push_back(candidates[i]);
combinationSumCore(candidates,i,ret,temp,curSum,target);
temp.pop_back();
curSum-=candidates[i];
}
}
}
};
