[Leetcode] Combination Sum II_leetcode ccombination sum ii-优快云博客

本文链接：https://blog.youkuaiyun.com/a1342504010/article/details/38664029

本文介绍了一个算法问题，即给定一组候选数和目标数，找出所有可能的组合，使得组合中的数相加等于目标数。每个数只能用一次，且结果不能包含重复的组合。文章提供了详细的C++实现代码。

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题目：

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums toT.

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]

现在是一个数字只能使用一次，且结果不能有重复，跟上一题只要改一个地方。

vector<vector<int> > combinares;
bool combinationsum(vector<int> input, vector<int> candidates, int begin, int target)
{
	if(target<0) return true;
	if(target==0) 
		{
			combinares.push_back(candidates);
			return  true;
		}
	int flag=false;
	for(int i=begin; i<input.size(); i++)
	{
		if(i>begin &&input[i]==input[i-1] ) continue;
		candidates.push_back(input[i]);
		flag=combinationsum(input, candidates, i+1, target-input[i]);//要指到下一个元素，不能重复
		candidates.pop_back();
		if(flag) break;
	}
	return  false;
}

vector<vector<int> > combinationSum2(vector<int> &candidates, int target)
{
	if(candidates.size()<1)  return combinares;
	vector<int> candidate;
        sort(candidates.begin(), candidates.end());
	combinationsum(candidates, candidate, 0, target);

	return combinares;
}