LeetCode: Combination Sum II

最新推荐文章于 2019-03-06 12:45:47 发布

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最新推荐文章于 2019-03-06 12:45:47 发布

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分类专栏： LeetCode 文章标签： numbers combinations path include function

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本文介绍了一个使用深度优先搜索（DFS）解决组合求和问题的方法。该问题要求在给定候选数集合中找到所有唯一组合，使得这些组合的元素之和等于目标数。每个数在每个组合中只能使用一次。文章提供了详细的C++实现代码。

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#include<iostream>
#include<vector>
#include<map>
#include <algorithm>
using namespace::std;


#include <algorithm>

class Solution {
public:
    void getsum(vector<int> & cands, int target, int start, vector<vector<int> > &results, vector<int> path){
        if(target == 0){    // found the sum
            results.push_back(path);
            return;
        }
        if(start >= cands.size() || target < 0)   return; //reach the end of candidates
        if(cands[start] > target)
            //getsum(cands, target, start+1, results, path);  //skip this element
			return;
		else{
            // exclude cands[start] or any following elements of equal value
            int pre = -1;
            //while(cands[start+k] == cands[start]) k++;
            //getsum(cands,target,start+k,results, path);
            // include cands[start]
			for(int i = start; i < cands.size(); i++)
			{
				if(cands[i] != pre)
				{
					path.push_back(cands[i]);
					getsum(cands,target-cands[i],i+1,results,path);
					pre = path.back();
					path.pop_back();
				}

			}
        }
    }
    vector<vector<int> > combinationSum2(vector<int> &candidates, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int> > results;
        vector<int> path;

        sort(candidates.begin(),candidates.end());
        getsum(candidates,target,0,results,path);
        return results;
    }
};

int main()
{
	vector<int> input;
	input.push_back(1);
	input.push_back(1);
	input.push_back(10);
	Solution ss;
	
	vector<vector<int> > result = ss.combinationSum2(input, 2);
	for(int i = 0; i < result.size(); i++)
	{
		for(int j = 0; j < result[i].size(); j++)
		{
			cout<<result[i][j];

		}
	cout<<endl;
	}
}

Solve this problem using DFS

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]

Round 2:

class Solution {
public:
    vector<vector<int> > combinationSum2(vector<int> &num, int target) {
        vector<vector<int> > result;
		if(num.empty())
			return result;
		vector<int> cur;
		std::sort(num.begin(), num.end());
		int sum = 0;
		dfs(num, target, result, cur, sum, 0);
		return result;
    }
private:
	void dfs(vector<int> &candidates, int target, vector<vector<int> > &result, vector<int> &cur, int &sum, int index)
		{
			if(sum > target)
				return;
			if(sum == target)
			{
				result.push_back(cur);
				return;
			}
			int pre = -1;
			for(int i = index; i < candidates.size(); i++)
			{
				if(sum + candidates[i] > target)
					break;
				if(candidates[i] == pre)
				    continue;
				cur.push_back(candidates[i]);
				sum += candidates[i];
				dfs(candidates, target, result, cur, sum, i+1);
				sum -= candidates[i];
				pre = cur.back();
				cur.pop_back();
			}
		}
};