(M)Backtracking:40. Combination Sum II
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这道题是Combination Sum的改进,给出一个数组,可能有重复数字,给出一个target,写出所有和为target的组合。这个题的变化是,一个数只能用一次。
首先给数组排序,然后按照上一个题的方法写就行。但是这样有一个问题,因为有重复数字,所以有些组合会加两遍。所以我用了set结构,最后再转化成vector返回。
class Solution {public: vector<vector<int>> combinationSum2(vector<int>& candidates, int target) { set<vector<int>> res; vector<int> r; sort(candidates.begin(), candidates.end()); backtracking(candidates, target, res, r, 0); vector<vector<int>> result; for(auto a : res) { result.push_back(a); } return result; } void backtracking(vector<int>& candidates, int target, set<vector<int>>& res, vector<int>& r, int level) { if(target == 0){ res.insert(r); return; } if(target < 0) return; if(level == candidates.size()) return; for(int i = level; i < candidates.size(); ++i) { if(candidates[i] <= target){ r.push_back(candidates[i]); backtracking(candidates, target-candidates[i], res, r, i+1); r.pop_back(); } } }};但其实也不用set这么麻烦,直接在for循环里面加一个while就可以,向上一个题一样。
class Solution {public: vector<vector<int>> combinationSum2(vector<int>& candidates, int target) { vector<vector<int>> res; vector<int> r; sort(candidates.begin(), candidates.end()); backtracking(candidates, target, res, r, 0); return res; } void backtracking(vector<int>& candidates, int target, vector<vector<int>>& res, vector<int>& r, int level) { if(target == 0){ res.push_back(r); return; } if(target < 0) return; if(level == candidates.size()) return; for(int i = level; i < candidates.size(); ++i) { if(candidates[i] <= target){ r.push_back(candidates[i]); backtracking(candidates, target-candidates[i], res, r, i+1); r.pop_back(); } while(i+1 < candidates.size() && candidates[i+1] == candidates[i]) ++i; } }};
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