LeetCode:39. Combination Sum(C++版本)
2016-02-28 09:25
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题目链接:
39. Combination Sum题目内容:
Given a set of candidate numbers (C) 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.
Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
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]
题目解释:
一个整型数组中存放着一些整数,从这些整数中挑选k个数,让这K个数的和为一个给定的数T,同一个数可以被挑选多次,求组成特定数T的不同方案。注意:
数组中的数和目标数T都是非负整数。
方案中的数必须按非递减顺序存放。
不能有重复的方案。
题目中给出了一个例子,一个候选数组为[2,3,6,7]要求组成特定的数7,可行方案为:[7],[2,2,3]
解题方案:
要求出所有的可行方案,我们就要去遍历所有方案,这个题目的解空间如下图所示:这其实是一个树结构的解空间,而且我们用到了一个剪枝函数:我们的方案要求不能重复,所以在挑选某数n时,比n小的数就不再遍历了,因为这个方案前面已经遍历过了,例如在遍历6时,我们只需在去遍历6,7这两种方案行不行即可,2和3我们不再去遍历。在这里我们用到了回溯的思想,当某一个数字不可行时,我们要回退到选择这个数字之前的状态。
AC代码:
class Solution { public: void combinate(vector<vector<int> > &res, vector<int> plan, vector<int> &candidates, int target) { //递归出口: 1、满足和。2、当前方案不可行。 if(target == 0) { res.push_back(plan); return ; } if(target < 0) return; for(int i = 0; i < candidates.size(); i++) { //剔除重复方案 if(candidates[i] < plan[plan.size() - 1]) continue; plan.push_back(candidates[i]); //递归 选择下一个 combinate(res, plan, candidates, target - candidates[i]); plan.pop_back(); } return ; } vector<vector<int> > combinationSum(vector<int>& candidates, int target) { vector<vector<int> > res; vector<int> plan; for(int i = 0; i < candidates.size(); i++) { plan.clear(); plan.push_back(candidates[i]); combinate(res, plan, candidates, target - candidates[i]); } return res; } };
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