[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]

LeetCode Source

思路：DFS。或者是采用回溯法。特别注意要先对原数组排序。

class Solution {
public:
vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
vector<vector<int> >ret;
vector<int> temp;
sort(candidates.begin(),candidates.end());
for(int i=0;i<candidates.size();++i){
dfs(ret,candidates,i,target,temp);
}
return ret;
}

void dfs(vector<vector<int> >&ret,vector<int> candidates,int i,int target,vector<int> temp){
if(candidates[i]==target){
temp.push_back(candidates[i]);
ret.push_back(temp);
return;
}
if(candidates[i]>target){
return;
}
if(candidates[i]<target){
temp.push_back(candidates[i]);
for(int j=0;j<candidates.size()-i;++j)
dfs(ret,candidates,i+j,target-candidates[i],temp);
}

}
};

AC了，但是发现运行时间长达435ms。分析后发现问题出在。

for(int j=0;j<candidates.size()-i;++j)
dfs(ret,candidates,i+j,target-candidates[i],temp);

我们对明显不会有解（j>target-candidates[i])的结果进行了DFS。修改如下，时间减少到30ms。

class Solution {
public:
vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
vector<vector<int> >ret;
vector<int> temp;
sort(candidates.begin(),candidates.end());
for(int i=0;i<candidates.size();++i){
if(candidates[i]<=target)
dfs(ret,candidates,i,target,temp);
}
return ret;
}

void dfs(vector<vector<int> >&ret,vector<int> candidates,int i,int target,vector<int> temp){
if(candidates[i]==target){
temp.push_back(candidates[i]);
ret.push_back(temp);
return;
}
if(candidates[i]<target){
temp.push_back(candidates[i]);
for(int j=0;j<candidates.size()-i;++j)
if(candidates[i+j]<=target-candidates[i]){
dfs(ret,candidates,i+j,target-candidates[i],temp);
}
}

}
};