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39. Combination Sum/组合总和

39. Combination Sum/组合总和

作者: 蜜糖_7474 | 来源:发表于2019-05-18 14:22 被阅读0次

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

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

Note:

All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
[7],
[2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
[2,2,2,2],
[2,3,3],
[3,5]
]

AC代码

void digui(vector<int>& candidates, int target, int idx, int sum, vector<int>& v, vector<vector<int>>& ans) {
    if (target == sum) {
        ans.push_back(v);
        return;
    }
    else if (target < sum) return;
    else {
        for (int i = idx; i < candidates.size(); ++i) {
            v.push_back(candidates[i]);
            digui(candidates, target, i, sum + candidates[i], v, ans);
            v.pop_back();
        }
    }
}

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<vector<int>> ans;
        vector<int> v;
        digui(candidates, target, 0, 0, v, ans);
        return ans;
    }
};

总结

应该算是深度优先搜索

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