LeetCode 39. Combination Sum 组合总和 (C++/Java)
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]
]
分析:
给定一个无重复元素的数组,要求用数组中的元素构成组合(可以重复),使元素和与target相同,且最后答案中不能有重复的组合。
利用深度优先搜索,在搜索的时候要传入一个索引,递归搜素时,从当前索引处向后搜索元素,防止最后有重复的元素组合,同时每搜索到一个元素,使当前目标修改为target-元素,当目标为0时,代表我们找到了一个解,加入到结果集合中。
可以先将所有元素由小到大排序,当搜索过程中当前元素大于目标值,此时后面的所有元素都不符合要求,提前退出循环。
程序:
C++
class Solution { public: vector<vector<int>> combinationSum(vector<int>& candidates, int target) { vector<vector<int>> res; vector<int> curr; sort(candidates.begin(), candidates.end()); dfs(res, curr, target, 0, candidates); return res; } void dfs(vector<vector<int>>& res, vector<int> curr, int target, int index, vector<int>& candidates){ if(target == 0){ res.push_back(curr); return; } for(int i = index; i < candidates.size(); ++i){ if(candidates[i] > target) break; //continue; curr.push_back(candidates[i]); dfs(res, curr, target-candidates[i], i, candidates); curr.pop_back(); } } };
Java
class Solution { public List<List<Integer>> combinationSum(int[] candidates, int target) { List<List<Integer>> res = new LinkedList<>(); LinkedList<Integer> curr = new LinkedList<>(); Arrays.sort(candidates); dfs(res, curr, target, 0, candidates); return res; } private void dfs(List<List<Integer>> res, LinkedList<Integer> curr, int target, int index, int[] candidates){ if(target == 0){ res.add(new LinkedList<>(curr)); return; } for(int i = index; i < candidates.length; ++i){ if(candidates[i] > target) break; curr.addLast(candidates[i]); dfs(res, curr, target-candidates[i], i, candidates); curr.removeLast(); } } }