【LeetCode】Combination Sum && II-优快云博客

本文链接：https://blog.youkuaiyun.com/snowwolf_yang/article/details/38933151

本文介绍了两种组合求和算法：CombinationSum与CombinationSumII。这两种算法分别用于寻找候选数集合中所有加和等于目标值的组合，区别在于是否允许重复使用集合中的元素。文章提供了详细的代码实现，并强调了组合的非递减排序及去重要求。

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Combination Sum

Total Accepted: 17172 Total Submissions: 64746 My Submissions

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

The same repeated number may be chosen from C unlimited number of times.

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 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3]

Combination Sum II

Total Accepted: 13526 Total Submissions: 55170 My Submissions

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:
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<vector<int> > ret;
        if (candidates.size() < 1) return ret;
        vector<int> tmp;
        combinationSum(ret, candidates, tmp, 0, target);
        return ret;
    }
    void combinationSum(vector<vector<int> > &ret, vector<int> &candidates, vector<int> &tmp, int index, int target) {
        if (target == 0) {
            ret.push_back(tmp);
            return;
        }
        if (target < 0) return;
        for (int i = index; i < candidates.size(); i++) {
            if (target >= candidates[i]) {
                tmp.push_back(candidates[i]);
                combinationSum(ret, candidates, tmp, i, target - candidates[i]);
                tmp.pop_back();
            }
        }
    }
};

class Solution {
public:
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        vector<vector<int> > ret;
        vector<int>  tmp;
        sort(candidates.begin(),candidates.end());
        combinationSumCore(ret,tmp,candidates,0,target);
        return ret;
    }
    void combinationSumCore(vector<vector<int> > &ret,vector<int>  &tmp,vector<int> &candidates,int index,int target) {
    	if(index == candidates.size() || target == 0) {
    		if(target == 0 && tmp.size() != 0) {
    			ret.push_back(tmp);
    			return;
    		}
    		return;
    	}
    	if(target-candidates[index] < 0) {
    		return;
    	}
    	tmp.push_back(candidates[index]);
    	combinationSumCore(ret,tmp,candidates,index,target-candidates[index]);
    	tmp.pop_back();
    	combinationSumCore(ret,tmp,candidates,index + 1,target);    
	}
};