【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, 0);
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 == 0)
{
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, 0);
return ret;
}
void Helper(vector<vector<int> >& ret, vector<int> cur, vector<int> &candidates, int target, int pos)
{
if(target == 0)
{
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();
}
}
}
};