LeetCode——Combination Sum

最新推荐文章于 2025-04-01 13:53:25 发布

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最新推荐文章于 2025-04-01 13:53:25 发布

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分类专栏： Online Judge 文章标签： leetcode

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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。算法采用递归搜索的方式实现，并确保了组合的独特性和非降序排列的要求。

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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]

» Solve this problem

递归搜索所有解

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<int>::iterator pos = unique(candidates.begin(), candidates.end());
        candidates.erase(pos, candidates.end());
        vector<vector<int> > ans;
        vector<int> record;
        searchAns(ans, record, candidates, target, 0);
        return ans;
    }
    
private:
    void searchAns(vector<vector<int> > &ans, vector<int> &record, vector<int> &candidates, int target, int idx) {
        if (target == 0) {
            ans.push_back(record);
            return;
        }
        if (idx == candidates.size() || candidates[idx] > target) {
            return;
        }
        for (int i = target / candidates[idx]; i >= 0; i--) {
            record.push_back(candidates[idx]);            
        }
        for (int i = target / candidates[idx]; i >= 0; i--) {
            record.pop_back();
            searchAns(ans, record, candidates, target - i * candidates[idx], idx + 1);            
        }
    }
};