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 (a, a, … ,a) must be in non-descending order. (ie, a ≤ a ≤ … ≤ a).
- 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：与Combination Sum不同之处在与每轮每个候选数字只能取一次，所以下次递归时考虑的当前元素的之后的元素
* 所以这次在index中存放的下标是当前元素的下一个位置，以便递归时直接跳过上次考察过的元素，避免重复考察。
* 但是在记录结果时需要回复记录在index中的下标。
* 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: 4 milli secs （基本在几毫秒）
* Progress: 22/22 test cases passed.
* (Judge Large)
* Run Status: Accepted!
* Program Runtime: 144 milli secs （基本稳定在一百四十几毫秒左右）
* Progress: 172/172 test cases passed.
*/
#include <vector>
#include <algorithm>
#include <functional>
#include <iostream>
using namespace std; class Solution {
private:
const int index_count;
vector<vector<int> > results;
public:
Solution() : index_count() {};
// 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 = ; i <= n; ++i)
{
result.push_back(candidates[index[i]-]); // 这里需要减一，因为下面每次记录索引时加了1
}
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+] = i+; // 记录当前考察的候选项索引并加一，下次考察是跳过上次考察过的元素，每轮每个元素值考察一次
backtrace(target, sum+candidates[i], candidates, index, n+);
}
}
vector<vector<int> > combinationSum2(vector<int> &candidates, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
sort(candidates.begin(), candidates.end()); int *index = new int[index_count];
memset(index, , 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, , candidates, index, ); delete[] index; // 去重
vector<vector<int> >::iterator end = results.end();
sort(results.begin(), end, less<vector<int> >());
vector<vector<int> >::iterator new_end = unique(results.begin(), results.end());
results.erase(new_end, end); 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.combinationSum2(candidates, target); for (size_t i = ; i < result.size(); ++i)
{
for (size_t j = ; j < result[i].size(); ++j)
{
cout << result[i][j] << ' ';
}
cout << endl;
}
cout << endl; return ;
}