LeetCode - Combination Sum

最新推荐文章于 2021-03-18 11:37:34 发布

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本文介绍了一种寻找候选数集合中所有唯一组合的算法，这些组合的和等于目标数。算法允许同一候选数无限次重复选取，并确保组合不重复且按非递减顺序排列。

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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, ..., a_k) must be in non-descending order. (ie, a₁ <= a₂ <=...<= a_k).
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]

class Solution {
public:
    void findPath(set<vector<int> >& numSet, vector<vector<int> >& dp,
		vector<int> &candidates, vector<int>& path,int value){
		if(value==0){
			vector<int> tmp(path);
			sort(tmp.begin(),tmp.end());
			numSet.insert(tmp);
			return;
		}
		for(int i=0;i<dp.size();i++){
			if(dp[i][value]!=-1){
				path.push_back(candidates[i]);
				findPath(numSet,dp,candidates,path,dp[i][value]);
				path.pop_back();
			}
		}
    }	 

    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        vector<vector<int> > result;
		if(candidates.empty()||candidates.size()==0){
			return result;
		}

		int n=candidates.size();  
		sort(candidates.begin(),candidates.end());

		vector<vector<int> > dp(n,vector<int>(target+1,-1));
		set<vector<int> > numSet;
		vector<bool> visited(target+1,false);
		vector<int> path;
		visited[0]=true; 

		for(int i=0;i<n;i++){
			if(candidates[i]>target){
				break;
			}
			for(int j=candidates[i];j<=target;j++){
				if(visited[j-candidates[i]]==true){
					dp[i][j]=j-candidates[i];
					visited[j]=true;
				}
			}
		}

		findPath(numSet,dp,candidates,path,target);
		result.assign(numSet.begin(),numSet.end());
        return result;
    }
};