leecode 解题总结：39. Combination Sum

#include <iostream>
#include <stdio.h>
#include <vector>
#include <algorithm>

using namespace std;
/*
问题:
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,

分析:
这是程序员面试金典的一道题目。求组成数的所有组合。应该是用递归做。
之前是n分的硬币用25,10,5,1的硬币表示分别枚举
i * denom <= total时，i从0到k的情况，然后递归做
2 3 6 7
此题应该也是用这种方式

输入:
4(数组元素个数) 7(目标值)
2 3 6 7
4 6
2 3 6 7
2 7
2 6
输出:
7,2 2 3
6,3 3,2 2 2
no result

关键:
1 前半部分求解结果，前半部分 与 后半部分结果 进行笛卡尔积 即为最终结果
int curCandidate = candidates.at(curCandidateIndex);
vector<vector<int>> results;//后半部分递归求解结果
vector<int> result;//前半部分求解结果，前半部分 与 后半部分结果 进行笛卡尔积 即为最终结果
for(int i = 0 ; i * curCandidate <= target ; i++)
{
result.clear();//清空上一次结果
for(int j = 0 ; j < i ; j++)
{
result.push_back(curCandidate);
}
results = combineSum(candidates , target -  i * curCandidate , curCandidateIndex + 1 );//得到多个结果需要和当前结果进行笛卡尔积拼接
int size = results.size();
//进行笛卡尔积拼接
for(int k = 0 ; k < size ; k++)
{
results.at(k).insert(results.at(k).end() , result.begin() , result.end());//插入到后面，前面是最小部分
totalResults.push_back(results.at(k));
}
//如果当前直接等于结果集i * curCandidate = target,直接压入结果中
if(i * curCandidate == target)
{
totalResults.push_back(result);
}
}
return totalResults;
*/

bool compare(int a, int b)
{
return a > b;
}

class Solution {
public:

vector<vector<int>> combineSum(vector<int>& candidates, int target , int curCandidateIndex)
{
vector< vector<int> > totalResults;
if(candidates.empty() || target <= 0 || curCandidateIndex < 0 || curCandidateIndex >= candidates.size())
{
return totalResults;
}
int curCandidate = candidates.at(curCandidateIndex);
vector<vector<int>> results;//后半部分递归求解结果
vector<int> result;//前半部分求解结果，前半部分 与 后半部分结果 进行笛卡尔积 即为最终结果
for(int i = 0 ; i * curCandidate <= target ; i++)
{
result.clear();//清空上一次结果
for(int j = 0 ; j < i ; j++)
{
result.push_back(curCandidate);
}
results = combineSum(candidates , target -  i * curCandidate , curCandidateIndex + 1 );//得到多个结果需要和当前结果进行笛卡尔积拼接
int size = results.size();
//进行笛卡尔积拼接
for(int k = 0 ; k < size ; k++)
{
results.at(k).insert(results.at(k).end() , result.begin() , result.end());//插入到后面，前面是最小部分
totalResults.push_back(results.at(k));
}
//如果当前直接等于结果集i * curCandidate = target,直接压入结果中
if(i * curCandidate == target)
{
totalResults.push_back(result);
}
}
return totalResults;
}

vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
//必须确保候选值从大到小排序
sort(candidates.begin() , candidates.end() , compare );
int curCandidateIndex = 0;
vector<vector<int>> results = combineSum(candidates, target , curCandidateIndex);
return results;
}
};

void print(vector<vector<int>>& results)
{
if(results.empty())
{
cout << "no result" << endl;
return;
}
int size = results.size();
for(int i = 0 ; i < size ; i++)
{
int len = results.at(i).size();
for(int j = 0 ; j < len ; j++)
{
cout << results.at(i).at(j) << " ";
}
cout << ",";
}
cout << endl;
}

void process()
{
int num;
vector<int> nums;
int target;
int value;
Solution solution;
vector< vector<int> > results;
while(cin >> num >> target)
{
nums.clear();
for(int i = 0 ; i < num ; i++)
{
cin >> value;
nums.push_back(value);
}
results = solution.combinationSum(nums , target);
print(results);
}
}

int main(int argc , char* argv[])
{
process();
getchar();
return 0;
}