combination-sum

（i）
【题目描述】Given a set of candidate numbers ( C ) and a target number ( T ), find all unique combinations in C where the candidate numbers sums to T .
The same repeated number may be chosen from C unlimited number of times.
Note:
All numbers (including target) will be positive integers.
Elements in a combination (a 1, a 2, … , a k) must be in non-descending order. (ie, a 1 ≤ a 2 ≤ … ≤ a k).
The solution set must not contain duplicate combinations.
For example, given candidate set2,3,6,7and target7,
A solution set is:
[7]
[2, 2, 3]

【解题思路】dfs

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class Solution {
private:
    vector<vector<int>> ans;
    vector<int> rs;

public:

    void dfs(int nowIndex,int nowSum, vector<int> cand, int candSize, int target)
    {
        if (nowSum > target)
            return;
        if (nowSum == target)
        {
            ans.push_back(rs);
            return;
        }
        for (int i = nowIndex; i < candSize; i++){
            if (nowSum + cand[i] <= target) // 因为cand是排序好的，显然会有大量的重复，应该避免，相当于剪枝
            {
                rs.push_back(cand[i]);
                dfs(i, nowSum + cand[i], cand, candSize, target);
                rs.pop_back();// 回溯
            }
            else
                continue;
        }
    }

    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {

        int len = candidates.size();
        if (len == 0)
            return ans;

        sort(candidates.begin(), candidates.end());// 排序

        dfs(0,0, candidates,len, target);

        return ans;
    }
};

（ii）
【题目描述】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 (a 1, a 2, … , a k) must be 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]

【解题思路】在上一题的基础上，保证数组中的每个数字不能够重复计算，所以需要设计去重的功能，在dfs时做了修改，发现仍然重复，所以有用到了c++stl map结构去重

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class Solution {
private:
    map<vector<int>, int> mp;
    vector<vector<int>> ans;
    vector<int> rs;
public:

    void dfs(int nowIndex,int nowSum, vector<int> cand, int candSize, int target)
    {
        if (nowSum > target)
            return;
        if (nowSum == target)
        {
            // 因为仍然会有重复，利用map/set去重
            if (mp[rs] == 0)
            {
                ans.push_back(rs);
                mp[rs] = 1;
            }
            return;
        }
        // 从当前的下一个开始，不包括自己，因为每个数组中的数字
        for (int i = nowIndex + 1; i < candSize; i++){
            if (nowSum + cand[i] <= target) // 因为cand是排序好的，显然会有大量的重复，应该避免，相当于剪枝
            {
                rs.push_back(cand[i]);
                dfs(i, nowSum + cand[i], cand, candSize, target);
                rs.pop_back();// 回溯
            }
            else
                continue;
        }
    }

    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {

        int len = candidates.size();
        if (len == 0)
            return ans;

        sort(candidates.begin(), candidates.end());// 排序

        dfs(-1,0, candidates,len, target);

        return ans;
    }
};