leetcode[40]Combination Sum II

最新推荐文章于 2024-11-20 10:20:22 发布

转载最新推荐文章于 2024-11-20 10:20:22 发布 · 44 阅读

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原文链接：http://www.cnblogs.com/Vae1990Silence/p/4283581.html

本文介绍了一种使用C++解决独特数字组合问题的方法，目标是找到给定候选数字集合中所有可能的组合，使得这些组合的总和等于特定的目标数。该方法强调了在解决方案集中不包含重复组合的重要性，并通过实例展示了如何实现这一过程。

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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, … , 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 searchTarget(vector<vector<int>> &res, vector<int> temp, vector<int> &candidates, int target, int index)
{
    if (target==0)res.push_back(temp);
    else
    {
        for (int i=index;i<candidates.size();i++)
        {
            if(i!=index&&candidates[i]==candidates[i-1])
                continue;
            if (target>=candidates[i])
            {
                temp.push_back(candidates[i]);
                searchTarget(res,temp,candidates,target-candidates[i],i+1);
                temp.pop_back();
            }
        }
    }
}
vector<vector<int>> combinationSum2(vector<int> &num, int target) 
{
    vector<vector<int>> res;
    vector<int> temp;
    sort(num.begin(),num.end());
    searchTarget(res,temp,num,target,0);
    return res;
}
};

转载于:https://www.cnblogs.com/Vae1990Silence/p/4283581.html