LeetCode 39: Combination Sum

最新推荐文章于 2020-02-24 23:26:48 发布

claviclelau

最新推荐文章于 2020-02-24 23:26:48 发布

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本文介绍了一种寻找候选数集合中所有可能组合的算法，这些组合的和等于目标数。文章详细阐述了递归深度优先搜索（DFS）的实现方法，并通过示例展示了如何找到所有非重复的组合。

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Difficulty: 3

Frequency: 3

Problem:

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

Solution:

class Solution {
public:
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        sort(candidates.begin(), candidates.end());
        vector<vector<int> > answer;
        vector<int> combination;
        DFS(answer, combination, 0, candidates, target, 0);
        return answer;
    }
    void DFS(vector<vector<int> > & answer, vector<int> & combination, int sum, vector<int> & candidates, int target, int order)
    {
        for (int i = order; i<candidates.size(); i++)
        {
            int temp_sum = sum + candidates[i];
            if (temp_sum == target)
            {
                answer.push_back(combination);
                answer[answer.size()-1].push_back(candidates[i]);
            }
            else if (temp_sum < target)
            {
                combination.push_back(candidates[i]);
                DFS(answer, combination, temp_sum, candidates, target, i);
                combination.pop_back();
            }
        }
    }
};

Notes:

Typical DFS.