[Leetcode] Combination Sum II

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

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

10,1,2,7,6,1,5

and target

8

,
A solution set is:

[1, 7]

[1, 2, 5]

[2, 6]

[1, 1, 6]

记录当前idx之前的元素在结果中是否被使用，如果之前的与当前元素相等的元素没有被使用的话，那么这个元素也不应该被使用，这样就可以去重了。比如[1, 1, 2]，如果第一个1没有被使用，那么第二个1也不能使用。

class Solution {
public:
void findNext(vector<int> &num, int target, vector<vector<int> > &res, vector<int> &v, int idx, int sum, vector<bool> &flag) {
if (sum > target || idx > num.size()) return;
if (sum == target) {
res.push_back(v);
return;
}
bool fflag = false;
for (int i = idx-1; i >= 0; --i) {
if (num[i] == num[idx] && !flag[i]) {
fflag = true;
break;
};
if (num[i] != num[idx]) break;
}
if (!fflag) {
v.push_back(num[idx]);
flag[idx] = true;
findNext(num, target, res, v, idx+1, sum+num[idx], flag);
flag[idx] = false;
v.pop_back();
}
findNext(num, target, res, v, idx+1, sum, flag);
}

vector<vector<int> > combinationSum2(vector<int> &num, int target) {
vector<vector<int> > res;
vector<bool> flag(num.size(), false);
vector<int> v;
sort(num.begin(), num.end());
findNext(num, target, res, v, 0, 0, flag);
return res;
}
};

直接找的话会有重复的答案，可以先把结果存在一个set里，最然把结果从set转存到vector里。代码如下。

class Solution {
public:
void findNext(vector<int> &num, int target, set<vector<int> > &res, vector<int> &v, int idx, int sum) {
if (sum > target || idx > num.size()) return;
if (sum == target) {
res.insert(v);
return;
}
v.push_back(num[idx]);
findNext(num, target, res, v, idx+1, sum+num[idx]);
v.pop_back();
findNext(num, target, res, v, idx+1, sum);
}

vector<vector<int> > combinationSum2(vector<int> &num, int target) {
set<vector<int> > res;
vector<int> v;
sort(num.begin(), num.end());
findNext(num, target, res, v, 0, 0);
vector<vector<int> > ret;
for (set<vector<int> >::iterator i = res.begin(); i != res.end(); ++i) {
ret.push_back(*i);
}
return ret;
}
};