题目描述(Medium)
Given a collection of candidate numbers (candidates
) and a target number (target
), find all unique combinations in candidates
where the candidate numbers sums to target
.
Each number in candidates
may only be used once in the combination.
Note:
- All numbers (including
target
) will be positive integers. - The solution set must not contain duplicate combinations.
题目链接
https://leetcode.com/problems/combination-sum-ii/description/
Example 1:
Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]
Example 2:
Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
[1,2,2],
[5]
]
算法分析
参考【LeetCode】106.Combination Sum,本题需要考虑重复数字,先对原始数组排序,并用一个变量记录上一个使用的数值,重复即跳过。
提交代码:
class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
vector<int> solution;
sort(candidates.begin(), candidates.end());
dfs(candidates, solution, target, 0, 0);int previous = -1;
return this->result;
}
private:
vector<vector<int>> result;
void dfs(vector<int>& candidates, vector<int>& solution, int target,
int cur_sum, int start) {
if (cur_sum == target && start <= candidates.size()) {
this->result.push_back(solution);
return;
}
int previous = -1;
for (int i = start; i < candidates.size(); ++i) {
cur_sum += candidates[i];
solution.push_back(candidates[i]);
if (cur_sum <= target && previous != candidates[i])
{
dfs(candidates, solution, target, cur_sum, i + 1);
previous = candidates[i];
}
cur_sum -= candidates[i];
solution.pop_back();
}
}
};