Leetcode 40

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ld;

#define REP(x, l, u) for(ll x = l;x<u;x++)
#define RREP(x, l, u) for(ll x = l;x>=u;x--)
#define se second
#define fi first
#define input(s) getline(cin,s)
typedef vector<int> VI;
typedef unordered_map<int,int> UMAPI;
const ll mod = 1e9 + 7;
#define len(x) x.size()

class Solution {
    vector<VI> ans;
    VI tmp;
    void search(VI& candidates, int rem, int ind){
        if(rem == 0){
            ans.emplace_back(tmp);
            return;
        }

        REP(i, ind, len(candidates)){
            if(i != ind && candidates[i] == candidates[i-1]) continue;
            if(candidates[i] <= rem){
                tmp.emplace_back(candidates[i]);
                search(candidates, rem - candidates[i], i + 1);
                tmp.pop_back();
            }
            else break;
        }
    }
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        search(candidates, target, 0);
        return ans;
    }
};

思路上基本和 39 一样，排序然后搜索。只需要注意同一种元素打头的搜索只需要一次就行了。比如[2, 2, 3]，当遍历并深搜这个数组的时候，遍历到第二个 2 的时候跳过就行了。