LeetCode:Combination Sum

Given a set of candidate numbers (C) (without duplicates) 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]

]

思路：

这个问题我看到就想到了找余下的值，也就是7-2=5，则找5在不在数组，不在就把2推进数组，然后5-2=3，找3,在的话就把3推进数组，然后就把这个小数组加入到大数组里，不过目前代码存在的问题是并不能找出所有解，只能找出部分，正在解决中orz。。。

目前出现的问题：

代码如下：
class Solution {
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
vector<vector<int>>res;
int flag = 0;
//bool is_add=false;
//int count=candidates.size();
for (int i = 0; i<candidates.size(); i++)
{
int temp = target;
vector<int>m_vec;
m_vec.push_back(candidates[i]);
while (temp>0)
{
if(target%candidates[i]==0&&target!=candidates[i])
{
int count=target/candidates[i];
while(count!=1)
{
m_vec.push_back(candidates[i]);
count--;
}
res.push_back(m_vec);
break;
}
temp = temp - candidates[i];
if (bfind(candidates, temp))
{
m_vec.push_back(temp);
temp = 0;
res.push_back(m_vec);
break;
}
if (temp == 0)
{
res.push_back(m_vec);
break;
}
m_vec.push_back(candidates[i]);
}
//j++;
}
return res;
}

bool bfind(vector<int>& candidates, int target)//二分查找
{
vector<int>test(candidates);
sort(test.begin(), test.end());
int div = (test.size()-1) / 2;
while (test[div] != target)
{
if (test[div]>target)
div = (div-1)/2;
else
div = (test.size() + div ) / 2   ;
if (test[div] == target)
return true;
else if (div == 0 && test[div] != target)
return false;
else if (div == test.size() - 1 && test[div] != target)
return false;
}
return true;
}
};