Leetcode-39 Combination Sum

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Combination Sum

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

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

Example 2

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

解题思路

回溯法。对于每一层，若 target<0，表示该组合不可行，直接返回上一层。若 target==0，说明得到可行组合，将其加入结果集并返回上一层。若 target>0，将其减去当前元素，并递归求解子问题。为避免加入重复组合，在求解子问题时不考虑之前元素。

复杂度分析

略（不会）。

代码

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class Solution {
	vector<vector<int>> res;
public:
	vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
		vector<int> sol;
		combination(candidates, target, sol, 0);
		return res;
	}

	void combination(vector<int> &candidates, int target, vector<int> &sol, int k) {
		if (target < 0) return;
		if (target == 0) {
			res.push_back(sol);
			return;
		}
		for (int i = k; i < candidates.size(); ++i) {
			sol.push_back(candidates[i]);
			combination(candidates, target - candidates[i], sol, i);
			sol.pop_back();
		}
	}
};