[leetcode] Combination Sum and Combination SumII

weixin_34067049發表於2015-09-16
LeetCode

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Note:

  • All numbers (including target) will be positive integers.
  • Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
  • The solution set must not contain duplicate combinations.

For example, given candidate set 2,3,6,7 and target 7, 
A solution set is: 
[7] 
[2, 2, 3] 

class Solution {
private:
    vector<vector<int> > ivvec;
public:
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        vector<int> ivec;
        sort(candidates.begin(), candidates.end());
        combinationSum(candidates, 0, target, ivec);
        return ivvec;
    }
    
    void combinationSum(vector<int> &candidates, int beg, int target, vector<int> ivec)
    { 
        if (0 == target)
        {
            ivvec.push_back(ivec);
            return;
        } 
        for (int idx = beg; idx < candidates.size(); ++idx)
        {
            if (target - candidates[idx] < 0)
                break;
            ivec.push_back(candidates[idx]);
            combinationSum(candidates, idx, target - candidates[idx], ivec);
            ivec.pop_back();
        }
    }
};


Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

Note:

  • All numbers (including target) will be positive integers.
  • Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
  • The solution set must not contain duplicate combinations.

For example, given candidate set 10,1,2,7,6,1,5 and target 8, 
A solution set is: 
[1, 7] 
[1, 2, 5] 
[2, 6] 
[1, 1, 6] 

與上題差別不大。依然是用DFS，基本的問題在於怎樣去重。

能夠新增一個剪枝: 噹噹前元素跟前一個元素是同樣的時候。假設前一個元素被取了，那當前元素能夠被取，也能夠不取，反過來假設前一個元素沒有取。那我們這個以及之後的所以同樣元素都不能被取。

(採用flag的向量作為標記元素是否被選取)

class Solution {
private:
    vector<vector<int> > ivvec;
    vector<bool> flag;
public:
    vector<vector<int> > combinationSum2(vector<int> &num, int target) {
        vector<int> ivec;
        sort(num.begin(), num.end());
        flag.resize(num.size());
        for (int i = 0; i < flag.size(); ++i)
            flag[i] = false;
        combinationSum2(num, 0, ivec, target, 0);
        return ivvec;
    }
    
    void combinationSum2(vector<int> &num, int beg, vector<int> ivec, int target, int sum)
    {
        if (sum > target)
            return;
        if (sum == target)
        {
            ivvec.push_back(ivec);
            return;
        }
        
        for (int idx = beg; idx < num.size(); ++idx)
        {
            if (sum + num[idx] > target) continue;
            if (idx != 0 && num[idx] == num[idx - 1] && flag[idx - 1] == false)
                continue;
            flag[idx] = true;
            ivec.push_back(num[idx]);
            combinationSum2(num, idx + 1, ivec, target, sum + num[idx]);
            ivec.pop_back();
            flag[idx] = false;
        }
    }
};


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