LEETCODE: Combination Sum-优快云博客

本文链接：https://blog.youkuaiyun.com/anyicheng2015/article/details/41963845

本文介绍了一种寻找候选数字集合中所有唯一组合的方法，这些组合的数字之和等于目标数值。算法允许同一数字可以被重复选择，并确保组合不重复且按非降序排列。

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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, … , 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]

排好序，然后通过递归解决问题。在任何时候，当前的数和它之后的数都有可能是下一个数。

class Solution {
public:
    void combinationSumInternal(vector<int> &candidates, vector<int> currentnums, int current, 
                                int target, int start, int end, vector<vector<int> > &results) {
        if(current == target) 
        {
            results.push_back(currentnums);
            return;
        }
        
        if(current > target)
            return;
        
        for(int ii = start; ii < end; ii ++) {
            currentnums.push_back(candidates[ii]);
            combinationSumInternal(candidates, currentnums, current + candidates[ii], target, ii, end, results);
            currentnums.pop_back();
        }
        
    }
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        vector<vector<int> > results;
        vector<int> currentnums;
        sort(candidates.begin(), candidates.end());
        combinationSumInternal(candidates, currentnums, 0, target, 0, candidates.size(), results);
        return results;
    }
};