LeetCode:Combination Sum

最新推荐文章于 2018-09-21 15:46:40 发布

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最新推荐文章于 2018-09-21 15:46:40 发布

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分类专栏： LeetCode OJ LeetCode解题记录文章标签： LeetCode Array Backtracking

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本文探讨了组合总和问题，即给定一组候选数和目标数，找出所有可能的组合，使得候选数之和等于目标数。解决方案允许无限次重复使用候选数。文章提供了C++代码实现，并附带详细解释。

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Combination Sum

Total Accepted: 95000 Total Submissions: 302249 Difficulty: Medium

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.
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]
]

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c++ code:

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        
        vector<vector<int>> combs;
        vector<int> comb;
        combinationSum(combs,comb,candidates,0,target);
        return combs;
    }
    
    // 自定义函数
    void combinationSum(vector<vector<int>>& combs, vector<int>& comb, vector<int>& candidates, int start, int target) {
        if (target < 0) return;
        else if(target==0) {
            combs.push_back(comb);
            return;
        }
        for(int i=start;i<candidates.size();i++) {
            comb.push_back(candidates[i]);
            combinationSum(combs,comb,candidates,i,target-candidates[i]);// 这里不是i+1，因为可以重用i
            comb.pop_back();
        }
    }
    
};