40. Combination Sum II

Leetcode Array Backtracking

题目

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]
]

分析

40题和39题相比，主要差别在每个数字只能使用一次：在explore步骤时采用i+1参数。对于重复的结果，笨办法是使用set。

#include <algorithm>
class Solution {
public:
    void combinationSum2Helper(vector<int>& candidates, int target, set<vector<int>>& res, vector<int>& chosen, int position){
        if (target == 0){
        // base case
            res.insert(chosen);
        }else{
            for(int i=position; (i<candidates.size()) && (candidates[i]<=target);i++){
                //choose
                chosen.push_back(candidates[i]);

                //explore
                combinationSum2Helper(candidates, target-candidates[i], res, chosen, i+1);

                //unchoose
                chosen.pop_back();

            }
        }

    }



    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        set<vector<int>> res;
        vector<int> chosen;
        sort(candidates.begin(), candidates.end());
        combinationSum2Helper(candidates, target, res, chosen, 0);
        vector<vector<int>> res_final(res.begin(), res.end());
        return res_final;
    }
};

有一种更巧妙的办法去除重复，需要仔细分析什么时候会出现重复。其实只有一种情况，那就是在挑选下一个元素时，有前后元素相同。所以可以直接在choose步骤，增加一句

if (i&&num[i]==num[i-1]&&i>position) continue;  // check duplicate combination

完整的程序为

#include <algorithm>
class Solution {
public:
    void combinationSum2Helper(vector<int>& candidates, int target, vector<vector<int>>& res, vector<int>& chosen, int position){
        if (target == 0){
        // base case
            res.push_back(chosen);
        }else{
            for(int i=position; (i<candidates.size()) && (candidates[i]<=target);i++){
                if ((i>position) && (candidates[i]== candidates[i-1])){// 跳过重复
                    continue;
                } else{
                    //choose
                    chosen.push_back(candidates[i]);

                    //explore
                    combinationSum2Helper(candidates, target-candidates[i], res, chosen, i+1);

                    //unchoose
                    chosen.pop_back();
                }

            }
        }

    }



    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>> res;
        vector<int> chosen;
        sort(candidates.begin(), candidates.end());
        combinationSum2Helper(candidates, target, res, chosen, 0);
        return res;
    }
};