算法分析与设计——LeetCode:39. Combination Sum

题目

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

, 
A solution set is: 
[
[7],
[2, 2, 3]
]

class Solution {
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
}
};

思路

这是从题目的讨论中看到的方法，循环遍历，如[2, 3, 6, 7]，当i为0，放入一个2进入combination，再继续在[2, 3, 6, 7]里选择；当i为1时，则不再有2，只在[3, 6, 7]里选择，最终可以涵盖所有情况。

代码

class Solution {
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
sort(candidates.begin(), candidates.end());
vector<vector<int> > res;
vector<int> combination;
combinationSum(candidates, target, res, combination, 0);
return res;
}
void combinationSum(vector<int> &candidates, int target, vector<vector<int> > &res, vector<int> &combination, int head) {
if (target == 0) {
res.push_back(combination);
return;
}
for (int i = head; i < candidates.size()
4000
&& candidates[i] <= target; i++) {
combination.push_back(candidates[i]);
combinationSum(candidates, target - candidates[i], res, combination, i);
combination.pop_back();
}
}
};