leetcode15. 3Sum

medium程度题

题目:

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: 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]
]

菜鸡没想出来，哎。。参考的别人的解法，先排序，然后夹逼，去重

其实挺简单的，要多想多积累啊。。。。。。

类似的可以拓展为k SUM ,时间复杂度O(max(n㏒n,N^(k - 1)))

AC解:

class Solution 
{
public:
    vector<vector<int>> threeSum(vector<int>& nums) 
    {
        vector<vector<int>> result;
        int size = nums.size();
        if (size < 3)
            return result;
        
       sort(nums.begin(),nums.end());
       const int sum = 0;
       auto last = nums.end();
       
       for (auto i = nums.begin(); i < nums.end() - 2; i++)
       {
           auto j = i + 1;
           if (i > nums.begin() && *i == *(i - 1))
               continue;
           
           auto k = last - 1;
           while (j < k)
           {//去重
               if (*i + *j + *k > sum)
               {
                   k--;
                   while (*k == *(k + 1))
                       k--;     
               }else if (*i + *j + *k < sum)
               {
                   j++;
                   while (*j == *(j - 1))
                       j++;  
               }else
               {
                   result.push_back({*i,*j,*k});
                   j++;
                   k--;
                   while (*j == *(j - 1) && *k ==*(k + 1) && j < k)
                   {
                       j++; 
                       k--;
                   }
               }
               
           }
        }
       
        return result;
    }
};

类似的，4sum题:

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: 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]
]

AC解:

class Solution 
{
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) 
    {
        vector<vector<int>> result;
        int size = nums.size();
        if (size < 4)
            return result;
            
        auto first = nums.begin();
        auto last = nums.end();
        sort(nums.begin(),nums.end());
        
        for (auto i = first; i < last - 3; i++)
        {
            if (i > first && *i == *(i - 1))
                continue;
            for (auto j = i + 1; j < last - 2; j++)
            {
                if (j > i + 1 && *j == *(j - 1))
                    continue;
                auto k = j + 1;
                auto l = last - 1;
                
                while (k < l)
                {
                    if (*i + *j + *k + *l > target)
                    {
                        l--;
                        while (*l == *(l + 1))
                            l--;
                    }
                    else if (*i + *j + *k + *l < target)
                    {
                        k++;
                        while (*k == *(k - 1))
                            k++;
                    }
                    else 
                    {
                        result.push_back({*i,*j,*k,*l});
                        k++;
                        l--;
                        while (*k == *(k - 1) && *l == *(l + 1))
                        {
                            k++;
                            l--;
                        }
                    }
                }
            }
        }
        
        return result;
    }
};