LeetCode39 - Combination Sum
问题分析：
此问题同之前括号匹配问题一样，使用递归思想
Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
The same repeated number may be chosen from candidates unlimited number of times.
Note:
- All numbers (including
target) will be positive integers. - The solution set must not contain duplicate combinations.
Example :
Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
[7],
[2,2,3]
]运用递归思想，重复利用一个数字的并代入求和，直到它的和达到target 或者它超过了target， 然后在递归返回，继续找下一个数字。
根据官网示例，[2, 3, 6, 7] 找寻 targey = 7 的组合：
递归开始：
push 2 [2] sum = 2
push 2 [2, 2] sum = 4
push 2 [2, 2, 2] sum = 6
push 2 [2, 2, 2, 2] sum = 8
sum = 8 > 7 递归返回 [2, 2, 2, 2] pop 2 [2, 2, 2]
由于[2, 2, 2, 2]的sum > target，且原数组有序的，则下一次为[2, 2, 2, ?]必大于target，
即设置一个flag，用于高效递归：
sum = 8 > 7 flag = false 递归返回 [2, 2, 2, 2] pop 2 [2, 2, 2] flag == false break --> [2, 2]
push 3 [2, 2, 3] sum = 7
sum = 7 = 7 flag = false 递归返回 [2, 2, 3] pop 3 [2, 2] flag == false break --> [2]
...
...
...
...
代码如下:
class Solution {
public:
vector> res;
vector res_elem;
vector> combinationSum(vector& candidates, int target) {
if (candidates.size() == 0) return res;
addvectorres (0, 0, target, candidates);
return res;
}
void addvectorres (int start, int sum, int target, vector candidates)
{
bool flag;
if (sum == target)
{
res.push_back(res_elem);
flag = false;
return;
}
else if (sum > target)
{
flag = false;
return;
}
else
{
for (int i = start; i < candidates.size(); i++)
{
res_elem.push_back(candidates[i]);
addvectorres (i, sum + candidates[i], target, candidates);
res_elem.pop_back();
if (flag = false) break;
}
}
}
}; 学习知识：
递归思想