【LeetCode】38.Combination Sum

最新推荐文章于 2025-05-25 18:32:47 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，这些组合的元素之和等于目标值。通过递归方法实现，确保了解决方案的唯一性和有效性，并避免了重复组合。

Description：

Given a set of candidate numbers (C) (without duplicates) 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]
]

题目思考：

使用递归向下查找数组，每次找到一个，减去继续找到子集再进行拼接。

避免重复，新的数字必须不能在当前数字的前面。

Solution：

class Solution {
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
return combinationSum(candidates, target, 0);
}
vector<vector<int>> combinationSum(vector<int>& candidates, int target, int beg) {
vector<vector<int>> ret;
if (target <= 0) {return ret;}
for (int i = beg; i < candidates.size(); i++) {
int iv = candidates[i];
vector<int> one;
if (iv == target) {
one.push_back(iv);
ret.push_back(one);
}

vector<vector<int>> suf = combinationSum(candidates, target-iv, i);
for (int j = 0; j < suf.size(); j++) {
suf[j].push_back(iv);
ret.push_back(suf[j]);
}
}
return ret;
}
};