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:
For example, given candidate set 2,3,6,7 and
target 7,
A solution set is:
[7]
[2, 2, 3]
C中的同一数字能够拿多次。找到的组合不能反复。
思路：dfs
每一层的第i个节点有 n - i 个选择分支
递归深度：递归到总和大于等于T就能够返回了
复杂度：时间O(n!)。空间O(n)
感觉測试数据有问题，我用以下两个代码。对于[1,1],1这个输入。输出的结果各自是[[1],[1]]和[[1]]，但两个代码都 Accepted 了。我感觉第二个代码才是正确的，输出结果没反复。
//代码一
vector<vector<int> > res;
vector<int> _nums;
void dfs(int target, int start, vector<int> &path){
if(target == 0) res.push_back(path);
for(int i = start; i < _nums.size(); ++i){
if(target < _nums[i]) return ; //这里假设没剪枝的话会超时
path.push_back(_nums[i]);
dfs(target - _nums[i], i, path);
path.pop_back();
}
}
vector<vector<int> >combinationSum(vector<int> &nums, int target){
_nums = nums;
sort(_nums.begin(), _nums.end());
vector<int> path;
dfs(target, 0, path);
return res;
}
//代码二
vector<vector<int> > res;
vector<int> _nums;
void dfs(int target, int start, vector<int> &path){
if(target == 0) res.push_back(path);
int previous = -1;
for(int i = start; i < _nums.size(); ++i){
if(_nums[i] == previous) continue;
if(target < _nums[i]) return ; //这里假设没剪枝的话会超时
previous = _nums[i];
path.push_back(_nums[i]);
dfs(target - _nums[i], i, path);
path.pop_back();
}
}
vector<vector<int> >combinationSum(vector<int> &nums, int target){
_nums = nums;
sort(_nums.begin(), _nums.end());
vector<int> path;
dfs(target, 0, path);
return res;
}
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原文：http://www.cnblogs.com/gcczhongduan/p/4741834.html