【LeetCode】39. Combination Sum (2 solutions)

Combination Sum

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 (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 2,3,6,7 and target 7, 
A solution set is: 
[7] 
[2, 2, 3]

寻找target成员的过程中，如果candidates[i]是组成target的成员之一，那么寻找target-candidates[i]的子问题与原题就完全一致，因此是典型的递归。

参数列表中：result设为全局变量，用于记录所有可行的路径，因此使用引用(&)；curPath是每次递归栈中独立部分，因此使用拷贝复制

解法一：使用map去重

class Solution {
public:
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<vector<int> > ret;
        map<vector<int>, bool> m;
        vector<int> cur;
        Helper(ret, cur, candidates, target, m, );
        return ret;
    }
    void Helper(vector<vector<int> >& ret, vector<int> cur, vector<int> &candidates, int target, map<vector<int>, bool> &m, int ind)
    {
        if(target == )
        {
            if(m[cur] == false)
            {
                ret.push_back(cur);
                m[cur] = true;
            }
        }
        else
        {
            for(int i = ind; i < candidates.size() && candidates[i] <= target; i ++)
            {// for each candidate
                int val = candidates[i];
                cur.push_back(val);
                Helper(ret, cur, candidates, target-val, m, i); // duplication allowed
                cur.pop_back();
            }
        }
    }
};

解法二：

稍作分析可知，重复结果的原因在于candidates中的重复元素。

因为我们默认每个位置的元素可以重复多次，而不同位置的元素是不同的。

对candidates的排序及去重的目的就是防止结果的重复，比如7 --> 2,2,3/2,3,2/3,2,2

注：去重函数unique的用法

1、先排序，因为unique只会去掉连续元素中的重复元素

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

2、调用unique函数

vector<int>::iterator iter = unique(candidates.begin(), candidates.end());

执行完毕之后，返回的iter指向去重之后新数组的尾部，

例如：1,2,2,4,4,5

得到：1,2,4,5,?,?

^

iter

3、最后删除iter到end()之间的所有元素

candidates.erase(iter, candidates.end());

class Solution {
public:
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<int>::iterator iter = unique(candidates.begin(), candidates.end());
        candidates.erase(iter, candidates.end());

        vector<vector<int> > ret;
        vector<int> cur;
        Helper(ret, cur, candidates, target, );
        return ret;
    }
    void Helper(vector<vector<int> >& ret, vector<int> cur, vector<int> &candidates, int target, int pos)
    {
        if(target == )
        {
            ret.push_back(cur);
        }
        else
        {
            for(int i = pos; i < candidates.size() && candidates[i] <= target; i ++)
            {
                //candidates[i] included
                cur.push_back(candidates[i]);
                //next position is still i, to deal with duplicate situations
                Helper(ret, cur, candidates, target-candidates[i], i);
                //candidates[i] excluded
                cur.pop_back();
            }
        }
    }
};