Leetcode組合せ総和シリーズの詳細

39. Combination Sum
Medium
Given a set of candidate numbers ( candidates ) (without duplicates) and a target number ( target ), find all unique combinations in  candidates  where the candidate numbers sums to  target .
The same repeated number may be chosen from  candidates  unlimited number of times.
Note:

All numbers (including  target ) will be positive integers.

The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

このテーマには重複する数字はなく、各配列は重複して選択することができ、各数字は選択しないかを選択することができるので、これはサブセットの再帰構造と同じで、ステップソート、枝切り操作を加えることができます.
繰り返し選択できるので、再帰するときの層数はiです

class Solution {
public:
    vector> res;
    vector> combinationSum(vector& candidates, int target) {
        sort(candidates.begin(),candidates.end());
        vector temp;
        helper(candidates,target,temp,0);
        return res;
    }
    void helper(vector& candidates, int target, vector temp, int start){
        if(target<0){
            return;
        }
        if(target==0){
            res.push_back(temp);
        }
        for(int i=start;itarget) break;
            target-=candidates[i];
            temp.push_back(candidates[i]);
            helper(candidates, target, temp, i);
            target+=candidates[i];
            temp.pop_back();
        }
    }
};

40. Combination Sum II
Medium
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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]
]

このテーマの1つの要素は1回しか選択できないし、重複要素も存在するので、標準的な剪断スキームを除去しなければならない.並べ替え剪定を追加します.

class Solution {
public:
    vector> res;
    vector> combinationSum2(vector& candidates, int target) {
        sort(candidates.begin(),candidates.end());
        vector temp;
        helper(candidates,target,temp,0);
        return res;
    }
    void helper(vector& candidates, int target, vector temp, int start){
        if(target<0){
            return;
        }
        if(target==0){
            res.push_back(temp);
        }
        for(int i=start;itarget) break;
            if(i!=start && candidates[i]==candidates[i-1]) continue;
            target-=candidates[i];
            temp.push_back(candidates[i]);
            helper(candidates, target, temp, i+1);
            target+=candidates[i];
            temp.pop_back();
        }
    }
};

216. Combination Sum III
Medium
Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.
Note:

All numbers will be positive integers.

The solution set must not contain duplicate combinations.

Example 1:

Input: k = 3, n = 7
Output: [[1,2,4]]

Example 2:

Input: k = 3, n = 9
Output: [[1,2,6], [1,3,5], [2,3,4]]

この問題の数字は1から9しかなく、一度しか使えないし、数を指定して、枝を切って検索する必要があります.

class Solution {
public:
    int nums[10] = {1,2,3,4,5,6,7,8,9};
    vector> res;
    vector> combinationSum3(int k, int n) {
        vector temp;
        helper(k,n,0,0, temp);
        return res;
    }
    void helper(int k, int n,int start, int sum, vector temp){
        if(sum>n || temp.size()>k) return;
        if(sum==n && temp.size()==k){
            res.push_back(temp);
            return;
        }
        for(int i=start;i<9;i++){
            sum+=nums[i];
            temp.push_back(nums[i]);
            helper(k,n,i+1,sum,temp);
            sum-=nums[i];
            temp.pop_back();
        }
    }
};