3Sum&&4Sum

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

• Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
• The solution set must not contain duplicate triplets.
• For example, given array S = {-1 0 1 2 -1 -4},
•  A solution set is: (-1, 0, 1) (-1, -1, 2)

 

思路1：先对数组排序，之后三个变量i(0->num.size()-1),j(initial:i+1),k(initial:num.size()-1)，i 扫描数组，判定sum=(num[i]+num[j]+num[k]==0) ? OK:(sum>0 ? --k:++j)；

问题的关键在于细节的考虑，若去处一下程序中的三个if判断，则对于上述example，（-1，-1,2）将有两个输出，判断在于对重复情况的去除

code:

class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num) {
        vector<vector<int> >result;
        if(num.size()<3)
            return result;
        
        sort(num.begin(),num.end());
        
        int sum=0;
        vector<int>tmp;
        for(int i=0;i<num.size();++i)
        {
            if(i>0&&num[i]==num[i-1])
                continue;
                
            int j=i+1;
            int k=num.size()-1;
            while(j<k)
            {
                if(j>i+1&&num[j]==num[j-1])
                {
                    ++j;
                    continue;
                }
                    
                if(k<num.size()-1&&num[k]==num[k+1])
                {
                    --k;
                    continue;
                }
                
                sum=num[i]+num[j]+num[k];
                if(sum<0)
                    ++j;
                else if(sum>0)
                    --k;
                else
                {
                    tmp.push_back(num[i]);
                    tmp.push_back(num[j]);
                    tmp.push_back(num[k]);
                    result.push_back(tmp);
                    tmp.clear();
                    ++j;
                    --k;
                }
            }
        }
        
        return result;
    }
};

思路二：也可以使用STL中相关函数解决上面问题

prev:

template <class BidirectionalIterator>
  BidirectionalIterator prev (BidirectionalIterator it,
       typename iterator_traits<BidirectionalIterator>::difference_type n = 1);

Get iterator to previous element. Returns an iterator pointing to the element that it would be pointing to if advanced -n positions.

If it is a random-access iterator, the function uses just once operator+ or operator-. Otherwise, the function uses repeatedly the increase or decrease operator (operator++ or operator--) on the copied iterator until n elements have been advanced.

upper_bound(first,last,x):返回值是寻找到元素下一位置的迭代器；

code:

class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num) {
        vector<vector<int> >result;
        if(num.size()<3)
            return result;
        
        sort(num.begin(),num.end());
        
        auto last=num.end();
        for(auto a=num.begin();a<prev(last);a=upper_bound(a,prev(last),*a))
        {
            for(auto b=next(a);b<prev(last);b=upper_bound(b,prev(last),*b))
            {
                const int c=-(*a+*b);
                if(binary_search(next(b),last,c))
                    result.push_back(vector<int> {*a,*b,c});
            }
        }
        
        return result;
    }
};

 

4Sum:

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

• Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
• The solution set must not contain duplicate quadruplets.

    For example, given array S = {1 0 -1 0 -2 2}, and target = 0.

    A solution set is:
    (-1,  0, 0, 1)
    (-2, -1, 1, 2)
    (-2,  0, 0, 2)

使用3Sum思路一，可得到如下代码，其中的关键同样在于重复的判断；

code：

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        vector<vector<int> >result;
        
        sort(num.begin(),num.end());
        vector<int>tmp;
        for(int i=0;i<num.size();++i)
        {
            if(i>0&&num[i]==num[i-1])
                continue;
                
            for(int j=i+1;j<num.size();++j)
            {
                if(j>i+1&&num[j]==num[j-1])
                    continue;
                    
                int m=j+1;
                int n=num.size()-1;
                
                while(m<n)
                {
                    if(m>j+1&&num[m]==num[m-1])
                    {
                        ++m;
                        continue;
                    }
                    if(n<num.size()-1&&num[n]==num[n+1])
                    {
                         --n;
                         continue;
                    }  
                        
                    int sum=num[i]+num[j]+num[m]+num[n];
                    
                    if(sum>target)
                        --n;
                    else if(sum<target)
                        ++m;
                    else
                    {
                        tmp.push_back(num[i]);
                        tmp.push_back(num[j]);
                        tmp.push_back(num[m]);
                        tmp.push_back(num[n]);
                        result.push_back(tmp);
                        tmp.clear();
                        --n;
                        ++m;
                    }
                }
            }
        }
        
        return result;
    }
};