leetcode 39 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]

]

class Solution {
public:
typedef vector<int>::iterator It;

vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
vector<vector<int> > ret;
sort(candidates.begin(), candidates.end());
vector<int> ans;
combine(candidates.begin(), candidates.end(), target, ans, ret);
return ret;
}
private:
void combine(It begin, It end, int target,
vector<int>& ans, vector<vector<int> >& ret){
for (It it = begin; it != end; ++it) {
vector<int> tmp(ans);
int val = *it;
if (val < target) {
tmp.push_back(val);
combine(it, end, target - val, tmp, ret);
} else if (val == target) {
tmp.push_back(val);
ret.push_back(tmp);
return; // 后面的都更大，所以返回
} else { // val > target
return;
}
}
}
}; // about 50%

参考后

快很多

class Solution {
public:
typedef vector<int>::iterator It;

vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
vector<vector<int> > ret;
sort(candidates.begin(), candidates.end());
vector<int> ans;
combine(candidates.begin(), candidates.end(), target, ans, ret);
return ret;
}
private:
void combine(It begin, It end, int target,
vector<int>& ans, vector<vector<int> >& ret){
for (It it = begin; it != end; ++it) {
int val = *it;
if (val < target) {
ans.push_back(val);
combine(it, end, target - val, ans, ret);
ans.pop_back();
} else if (val == target) {
ans
a490
.push_back(val);
ret.push_back(ans);
ans.pop_back();
return; // 后面的都更大，所以返回
} else { // val > target
return;
}
}
}
}; // about 75%