40. Combination Sum II

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.

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]
]

思路

和 39. Combination Sum 思路一致，不同的两点：

• Each number in candidates may only be used once in the combination.

• The solution set must not contain duplicate combinations.

第一点，每个数字只能使用一次，在递归时，下一个传参index加一即可；

第二点，必须包含不同的组合，首先想到的是用 std::set<vector<int>> 来存储返回值，但效率太低。另一种处理方式是在递归时增加判断条件，判断当前索引下的candidates值和前一个索引是否相同，相同则直接跳过该次递归：if (i > index && candidates.at(i) == candidates.at(i-1))

C++11

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>> res;
        std::sort(candidates.begin(), candidates.end());
        dfs(candidates, target, vector<int>{}, 0, res);
        return res;
    }

    void dfs(const vector<int> &candidates, int target, vector<int> curr, size_t index, vector<vector<int>> &res) {
        if (target == 0) {
            res.emplace_back(curr);
            return;
        }
        for (auto i = index; i < candidates.size(); ++i) {
            if (i > index && candidates.at(i) == candidates.at(i-1))
                continue;
            else if (candidates[i] <= target) {
                curr.emplace_back(candidates[i]);
                dfs(candidates, target - candidates[i], curr, i + 1, res);
                curr.pop_back();
            } else break; 
        }
    }
};