40. Combination Sum II

最新推荐文章于 2025-08-17 23:28:34 发布

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最新推荐文章于 2025-08-17 23:28:34 发布

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分类专栏： leetCode 文章标签： leetcode 递归

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本文介绍了一种给定候选数集合及目标数时，寻找所有唯一组合的算法实现。通过递归方法并确保每个数仅使用一次来找到总和等于目标数的组合，同时避免重复组合。

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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.
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]
]
思路：
    升序排序
    必须跳过重复的组合
	采用递归来寻找组合
	
*/

#include <stdio.h>
#include <iostream>
#include <string>
#include <sstream>
#include <vector>
#include <algorithm>
using namespace std;


class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target)
	{
        sort(candidates.begin(),candidates.end());
		vector<vector<int> >res;
		vector<int> combination;
		combinationS(res,candidates,combination,target,0);
		return res;
    }
private:
    void combinationS(vector<vector<int> > &res,vector<int> &candidates,vector<int> &combination,int target,int begin)
	{
		if(!target)
		{
			res.push_back(combination);
			return;
		}
		else{
			for(int i=begin;i<candidates.size() && candidates[i]<=target;i++)
			{
				if(i && candidates[i]==candidates[i-1] && i>begin) continue;//跳过重复部分
				combination.push_back(candidates[i]);
				combinationS(res,candidates,combination,target-candidates[i],i+1);
				combination.pop_back();
			}
		}
	}
};
int main()
{
    Solution mys;
	vector<std::vector<int> > res;
	vector<int> candidates;//={2,3,7};[10, 1, 2, 7, 6, 1, 5]
	candidates.push_back(1);
	candidates.push_back(1);
	candidates.push_back(2);
	candidates.push_back(7);
	candidates.push_back(6);
	candidates.push_back(1);
	candidates.push_back(5);
	int target=8;
	res=mys.combinationSum2(candidates,target);
	for(int i=0;i<res.size();i++)
	{
	    for(int j=0;j<res[i].size();j++)
		    cout<<res[i][j];
		cout<<endl;
	}
	
		
    return 0;	
}