***(leetcode_backtracking) Combination Sum II

Combination Sum II

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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] 

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 Array Backtracking

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回溯法，唯一注意的是，1,1,1  target=2的話，會產生重複數據。 

去重的判斷條件是 

1、如果i和i-1相等，則只有i-1加入到了求和的集合中才能將i算入 ，如果i-1沒加入，則說明之前某一組肯定把i-1節點加入了，這裏如果加入i節點，則會和加入i-1節點的重複！！！注意0節點就okay~

2、如果不相等，正常處理 

class Solution {
    vector<bool> flag;
    vector<vector<int> > ret;
    void dfs(vector<int> &num, int curDep, int maxDep, int sum, int target){
        if(sum==target){
            vector<int> ans;
            for(int i=0;i<num.size();i++)
                if(flag[i])
                    ans.push_back(num[i]);
            ret.push_back(ans);
            return;
        }
        if(sum>target||curDep==maxDep) 
            return;
        if((curDep==0)||(curDep!=0&&( (num[curDep]==num[curDep-1]&&flag[curDep-1]) || (num[curDep]!=num[curDep-1]) ) ) ){
            flag[curDep]=true;
            dfs(num,curDep+1, maxDep, sum+num[curDep], target);
        }
        flag[curDep]=false;
        dfs(num,curDep+1, maxDep, sum, target);
    }
public:
    vector<vector<int> > combinationSum2(vector<int> &num, int target) {
        sort(num.begin(),num.end());
        for(int i=0;i<num.size();i++)
            flag.push_back(false);
        dfs(num, 0, num.size(), 0, target);
        return ret;
        
    }
};