原题
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]
]
题解
讲义
考察知识点：
回溯法（深度优先搜索，Depth-First Search）
回溯法搜索轨迹示意图：
代码
[code lang=”cpp”] class Solution { public: vector<vector <int>> combinationSum(vector<int>& candidates, int target) { vector<vector <int>> result{}; vector<int> candidate{}; sort(candidates.begin(), candidates.end()); sum(result, candidates, target, candidate, 0); return result; } void sum(vector<vector <int>>& result, vector<int>& candidates, int target, vector</int><int>& candidate, int level) { // up to the target if (target == 0) { result.push_back(candidate); return; } for (int i = level; i < candidates.size() && candidates[i] <= target; ++i) { candidate.push_back(candidates[i]); sum(result, candidates, target – candidates[i], candidate, i); candidate.pop_back(); } } }; [/code]文章来源：胡小旭 => 小旭讲解 LeetCode 39. Combination Sum 回溯法
题解视频背景音乐：《Sad Angel》