Leetcode之Combination Sum

最新推荐文章于 2022-02-17 10:27:55 发布

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本文深入探讨了在给定一组候选数字和目标数字的情况下，如何寻找所有可能的组合，使得这些候选数字的总和等于目标数字。算法允许同一数字可以被无限次重复选择，并确保解决方案集中不包含重复的组合。

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题目：

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]
]

代码：

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

	for (int i = 0; i < candidates.size(); i++) {
		vector<vector<int>> v = combinationSum(candidates, target - candidates[i]);
		if (v.size() > 0) {
			for (int j = 0; j < v.size(); j++) {
				v[j].push_back(candidates[i]);
				sort(v[j].begin(), v[j].end());
				m.push_back(v[j]);
			}
		}
	}
    sort(m.begin(), m.end());
	m.erase(unique(m.begin(), m.end()), m.end());
	return m;
    }
};

注：

unique之前需要先sort