40. Combination Sum II**

40. Combination Sum II**

https://leetcode.com/problems/combination-sum-ii/

题目描述

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]
]

C++ 实现 1

DFS + Backtracing. 其实 39. Combination Sum** 中的代码可以直接用, 但是由于此题中每个元素只能被用一次, 所以 dfs 传入的是 i + 1 而不是 i; 另一方面, 如果 nums[i] == nums[i - 1], 那么 continue.

class Solution {
private:
    void dfs(const vector<int>& nums, int target, int start, vector<int> &cur, vector<vector<int>> &res) {
        if (target == 0) {
            res.push_back(cur);
            return;
        }
        for (int i = start; i < nums.size() && nums[i] <= target; ++ i) {
            if (i > start && nums[i] == nums[i - 1]) continue;
            cur.push_back(nums[i]);
            dfs(nums, target - nums[i], i + 1, cur, res);
            cur.pop_back();
        }
    }
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        std::sort(candidates.begin(), candidates.end());
        vector<vector<int>> res;
        vector<int> cur;
        dfs(candidates, target, 0, cur, res);
        return res;
    }
};