一. 题目描述
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 once number of times.
Note:
? All numbers (including target) will be positive integers.
? Elements in a combination (a1, a2, …, ak) must be in non-descending order. (ie, a1 > a2 > … > ak).
? The solution set must not contain duplicate combinations.
For example, given candidate set 10,1,2,7,6,1,5 and target 8, A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]
二. 题目分析
该题与之前的Combination Sum的解法类似,均可使用深度优先搜索来解。不同的是该题需要注意如何避免组合重复,因为不能重复,所以要跳过一样的数字。
例如:一个整数集合:[2 2 3],当我们要使用第二个2时,我们要检查他的前面一个2是否使用了,当未被使用时第二个2就不能使用;。
三. 示例代码
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
class Solution
{
public:
vector<vector<int> > combinationSum2(vector<int> &candidates, int target)
{
vector<int> temp; // 用于存放临时组合
sort(candidates.begin(), candidates.end());
combinationDFS(candidates, temp, 0, target);
return result;
}
private:
vector<vector<int> > result;
void combinationDFS(vector<int> &candidates, vector<int> &temp, size_t index, int target)
{
if (target == 0)
{
result.push_back(temp);
return;
}
else
{
int prev = -1;
for (size_t i = index; i < candidates.size(); ++i)
{
// 由于candidates中元素可能有重复,以下操作的意义是判断上轮循
// 环是否选择了candidates[i],是则跳过选择下一个candidates元素
// 直到下标到达比prev大的元素,选择该元素进行下一轮递归
if (prev == candidates[i])
continue;
if (candidates[i] > target)
return;
prev = candidates[i];
temp.push_back(candidates[i]);
combinationDFS(candidates, temp, i + 1, target - candidates[i]);
temp.pop_back();
}
}
}
};
四. 小结
相关题目参照:
时间: 2024-09-29 21:54:17