39. Combination Sum
题目描述:
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] ]题目链接:39. Combination Sum
算法描述:
根据题意,给定一个数组和一个目标值,结果返回能够产生这个目标值的组合,产生的方法为数组中数相加,且组合种类不可重复。
我们采用深度优先遍历算法来完成。首先找到一个数,如果小于目标值,则添加进 temp 容器,运用深度遍历继续带入函数……直到满足要求(目标值等于当前 temp 容器中数的和),或数的和大于目标值。当满足要求时,将满足要求的容器 temp 放入结果容器 ans,当 temp 中数的和大于目标值时,取出当前遍历中最新放入的数。
代码:
class Solution { public: vector<vector<int>> ans; vector<int> temp; vector<vector<int>> combinationSum(vector<int>& candidates, int target) { if(candidates.size()==0){ return ans; } function_dfs(0, 0, candidates, target); return ans; } void function_dfs(int pos, int sum, vector<int>& candidates, int target){ if(sum==target){ ans.push_back(temp); return; } else if(sum>target){ return; } else{ for(int i=pos; i<candidates.size(); i++){ temp.push_back(candidates[i]); function_dfs(i, sum+candidates[i], candidates, target); temp.pop_back(); } } } };