leetcode之Combination Sum

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] 

思路：典型的递归程序，对每次遍历，取当前数和不取当前数进行遍历。

class Solution {public:    void combinationSum(vector<int>& candidates,int index,int currentSum,int target,vector<int>& path)    {    if(currentSum == target)    {    res.push_back(path);    return;    }    int length = candidates.size();    if(index == length || currentSum > target)return;    path.push_back(candidates[index]);    combinationSum(candidates,index,currentSum+candidates[index],target,path);//重复添加当前节点    path.pop_back();    combinationSum(candidates,index+1,currentSum,target,path);//跳过当前节点    }        vector<vector<int> > combinationSum(vector<int> &candidates, int target) {        sort(candidates.begin(),candidates.end());    vector<int> path;    combinationSum(candidates,0,0,target,path);    return res;    }private:    vector<vector<int> > res;};

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 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] 

class Solution {public:    void combinationSum2(vector<int>& num,int index,int currentSum,int target,vector<int>& path)    {    if(currentSum == target)    {    hash.insert(path);//防止重复    return;    }    int length = num.size();    if(index == length || currentSum > target)return;    path.push_back(num[index]);    combinationSum2(num,index+1,currentSum+num[index],target,path);//取当前节点    path.pop_back();    combinationSum2(num,index+1,currentSum,target,path);//不取当前节点    }        vector<vector<int> > combinationSum2(vector<int> &num, int target) {    sort(num.begin(),num.end());    vector<int> path;    combinationSum2(num,0,0,target,path);    vector<vector<int> > res(hash.begin(),hash.end());    return res;    }private:    set<vector<int> > hash;};

如果是求总数，而且不需要去除重复，可以看这里