39. Combination Sum
problem description
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 1:
Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
[7],
[2,2,3]
]
Example 2:
Input: candidates = [2,3,5], target = 8,
A solution set is:
[
[2,2,2,2],
[2,3,3],
[3,5]
]
algorithm thought
组合数相加最后和为target,基本需要遍历所有集合。使用回溯法得到所有的集合,判断和是否为target。由于数字相加如果已经大于target,就没必要继续遍历了。所以加入了剪枝策略,首先sort数组,为了先遍历小数字后遍历大数字。然后在进入下一个函数之前判断是否满足剪枝策略,如果满足,就不进入函数。
code
class Solution {
public:
std::vector<std::vector<int> > combinationSum(std::vector<int> &candidates, int target) {
std::sort(candidates.begin(), candidates.end());
std::vector<std::vector<int> > res;
std::vector<int> combination;
combinationSum(candidates, target, res, combination, 0);
return res;
}
private:
void combinationSum(std::vector<int> &candidates, int target, std::vector<std::vector<int> > &res, std::vector<int> &combination, int begin) {
if (!target) {
res.push_back(combination);
return;
}
for (int i = begin; i != candidates.size() && target >= candidates[i]; ++i)
if (i == begin || candidates[i] != candidates[i - 1]) {
combination.push_back(candidates[i]);
combinationSum(candidates, target - candidates[i], res, combination, i);
combination.pop_back();
}
}
};
algorithm analysis
回溯法世家复杂度是O(2^n),很慢,但是这里加入了剪枝策略,还是一定程度上加快了运行速度的。
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