题目如下:
My Submissions Question Solution
Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
Each number in C may only be used once in the combination.
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 10,1,2,7,6,1,5 and target 8,
A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]
分析如下:
在前面一题 Combination Sum的基础上,增加滤重的一句话,使得题目条件得到满足。注意对题目条件的理解。 Each number in C may only be used once in the combination. 这句话的意思是,每个number可以不被使用,也可以使用,如果使用,只能用一次。所以,如果number a 出现了n次,那么在结果得到的组合中,a可以被用 0 ~ n次。
我一开始写了下面代码快中的第①句进行滤重,但是结果非常诡异,输出了不在输入中的元素,猜测是数组读取越界了,也就是i越界,所以增加了第②句控制i的范围。
其实把第①和第②简化一下,就成了第③。
我的代码:
//NOTE: Each number in C may only be used once in the combination. 这句话的意思是,每个number可以不被使用,也可以使用,如果使用,只能用一次。所以,如果number a 出现了n次,那么在结果得到的组合中,a可以被用 0 ~ n次。 class Solution { public: void combinationSumMy(vector<int> &candidates, vector<int> &every, vector<vector<int> > & final, int target, int start) { if (target == 0) { final.push_back(every); return; }else { for (int i = start; i < candidates.size(); ++i) { //while ((i > start) && (candidates[i] == candidates[i - 1])) i++; //① 滤重 //if (i >= candidates.size() ) return; //② 容易忘写 if (i > start && candidates[i] == candidates[i - 1]) continue; //③ 滤重 用本句来代替上面两句,更简洁。 if (candidates[i] > target) return; every.push_back(candidates[i]); combinationSumMy(candidates, every, final, target - candidates[i], i+1); every.pop_back(); } } } vector<vector<int> > combinationSum2(vector<int> &num, int target) { std::sort(num.begin(), num.end()); vector<int> every; vector<vector<int> > final; every.clear(); final.clear(); combinationSumMy(num, every, final, target, 0); return final; } };