Combination Sum

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, … ,a_k) must be in non-descending order. (ie, a₁ ≤ a₂ ≤ … ≤ a_k).
• 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] 

/** 整理的提交代码 * 处理复杂度为 * 主要思路：回溯法http://www.leetcode.com/2010/09/print-all-combinations-of-number-as-sum.html * 提交结果：  * (Judge Small)  * Run Status: Accepted!  * Program Runtime: 8 milli secs （基本在几毫秒） * Progress: 17/17 test cases passed.  * (Judge Large)  * Run Status: Accepted! * Program Runtime: 148 milli secs（基本稳定在一百四十几毫秒左右） * Progress: 168/168 test cases passed. */#include <vector>#include <algorithm>#include <iostream>using namespace std;class Solution {private:const int index_count;vector<vector<int> > results;public:Solution() : index_count(10000) {};// index记录当前找到的候选数字，n表示当前正在找第几个，n是index的下标不是candidates的下标void backtrace(int target, int sum, vector<int> &candidates, int index[], int n){if (sum > target){return;// 回溯}if (sum == target){vector<int> result;for (int i = 1; i <= n; ++i){result.push_back(candidates[index[i]]);}results.push_back(result);return;// 此处可以不加，如果不加return由于都是正整数，到下面的计算时会多进行一次无用的递归。}// 深度搜索，为了避免重复，每次从当前候选项索引到结尾，上面的i=index[n]可以看出for (int i = index[n]; i < candidates.size(); ++i){index[n+1] = i;// 记录当前考察的候选项索引backtrace(target, sum+candidates[i], candidates, index, n+1);}}vector<vector<int> > combinationSum(vector<int> &candidates, int target) {// Start typing your C/C++ solution below// DO NOT write int main() functionsort(candidates.begin(), candidates.end());int *index = new int[index_count];memset(index, 0, sizeof(int)*index_count);results.clear();// 提交到leetcode的测试系统上必须添加，它应该是使用一个对象测试所有测试用例。backtrace(target, 0, candidates, index, 0);delete[] index;return results;}};int main(){vector<int> candidates;int number;cout << "input candidates: ";while (cin >> number){candidates.push_back(number);}// 清除缓冲区cin.sync();cin.clear();int target;cout << "input target: ";cin >> target;vector<vector<int> > result;Solution s;result = s.combinationSum(candidates, target);for (size_t i = 0; i < result.size(); ++i){for (size_t j = 0; j < result[i].size(); ++j){cout << result[i][j] << ' ';}cout << endl;}cout << endl;return 0;}