LeetCode - Combination Sum II

最新推荐文章于 2021-03-19 10:57:05 发布

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本文介绍了一个算法问题：给定一组候选数字和目标数字，找出所有可能的组合，使得候选数字之和等于目标数字。该算法确保了组合的独特性和非递减顺序，并避免了重复的组合。

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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,..., 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 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:
    map<int,int> numMap;
	bool check(vector<int>& tmp){
		map<int,int>::iterator iter;
		int curr=tmp[0],num=1;
		for(int i=1;i<tmp.size();i++){
			if(tmp[i]==curr){
				num++;
			}else{
				iter=numMap.find(curr);
				if(num>iter->second){
					return false;
				}
				curr=tmp[i];
				num=1;
			}
		}
        iter=numMap.find(curr);
    	if(num>iter->second){
				return false;
		}
		return true;
	}

    void findPath(set<vector<int> >& numSet, vector<vector<int> >& dp,
			vector<int> &num, vector<int>& path,int value){
		if(value==0){
			vector<int> tmp(path);
			sort(tmp.begin(),tmp.end());
			if(check(tmp)){
				numSet.insert(tmp);
			}
			return;
		}
		for(int i=0;i<dp.size();i++){
			if(dp[i][value]!=-1){
				path.push_back(num[i]);
				findPath(numSet,dp,num,path,dp[i][value]);
				path.pop_back();
			}
		}
    }	
	
    vector<vector<int> > combinationSum2(vector<int> &num, int target) {
        vector<vector<int> > result;
		if(num.empty()||num.size()==0){
			return result;
		}
		int n=num.size();  
		sort(num.begin(),num.end());
	    numMap.clear();
		map<int,int>::iterator iter;
		for(int i=0;i<n;i++){
			iter=numMap.find(num[i]);
			if(iter==numMap.end()){
				numMap.insert(make_pair(num[i],1));
			}else{
				iter->second++;
			}
		}
		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(num[i]>target){
				break;
			}
			for(int j=target;j>=num[i];j--){
				if(visited[j-num[i]]==true){
					dp[i][j]=j-num[i];
					visited[j]=true;
				}
			}
		}

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