leetcode 040 —— 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]

 

思路：依旧回溯法，但是要加一个判定重复的步骤

class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
vector<vector<int>> res;
vector<int> path;
sort(candidates.begin(), candidates.end());

int sum = 0;
scan(0, sum, target, path, candidates, res);

return res;

}
void scan(int level, int &sum, int &target,
vector<int> &path, vector<int>& candidates, vector<vector<int>>& res){
if (sum > target)
return;
if (sum == target){
res.push_back(path);
return;
}
for (int i = level; i < candidates.size(); i++){
sum += candidates[i];
path.push_back(candidates[i]);
scan(i + 1, sum, target, path, candidates, res);
sum -= candidates[i];
path.pop_back();
while (i < candidates.size()-1 && candidates[i] == candidates[i + 1]) //一定要加，因为去出去相同的答案
i++;
}
}
}a;