[Leetcode] Combination Sum II

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 is used once in the combination.

Note:

• All numbers (including target) would be positive integers.
• elements in a combination (a 1 ,  a 2 , ...,  a k ) must is in non-descending order. (Ie, a 1  ≤ a 2  ≤ ... ≤ a k ).
• The solution set must not contain duplicate combinations.

For example, given candidate set10,1,2,7,6,1,5and target8,
A Solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]

Problem Solving Ideas:

Similar to combination sum, the difference is that each candidate element can only be used once.

Class Solution {public:vector<vector<int>> combinationSum2 (vector<int>& candidates, int target        ) {vector<vector<int>> result;        int len=candidates.size ();        if (len==0) {return result;        } std:sort (Candidates.begin (), Candidates.end ());  Map<int, int> Keytonumber;        The equivalent coefficient, which indicates how many times each number appears, notice that it becomes subscript, not number.  Set<vector<int>> contains;    Whether it contains this set of solutions GetResult (result, contains, candidates, 0, Keytonumber, target); } void GetResult (vector<vector<int>>& result, set<vector<int>>& contains, vector&lt             ;int>& uniquecandidates, int candidateindex, map<int, int>& keytonumber, int left) {if (left<0) {        Return            } if (left==0) {vector<int> item; For (Map<int, Int>::iterator it=keytonumber.begin (); It!=keytonumber.end (); it++) {int number=it->s         Econd;       int key = it->first;                for (int i=0; i<number; i++) {item.push_back (Uniquecandidates[key]);                }} if (Contains.find (item) ==contains.end ()) {Result.push_back (item);            Contains.insert (item);        } return;        } if (Candidateindex>=uniquecandidates.size ()) {return;        } int number=0; while (LEFT&GT;=0&AMP;&AMP;NUMBER&LT;2) {//used 0 times or once if (number!=0) keytonumber[candidateindex]=num            ber            GetResult (result, contains, uniquecandidates, Candidateindex+1, Keytonumber, left);            if (number!=0) {keytonumber.erase (candidateindex);            } left = Left-uniquecandidates[candidateindex];        number++; }    }};

[Leetcode] Combination Sum II