18. 4Sum

目录

• 题目描述
• 方法1：set去重
  • 代码
• 方法2：手动进行去重复处理
  • 代码

题目描述

Given an arraySof 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)

方法1：set去重

代码

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &nums, int target)
    {

    	set<vector<int>> ret;
        sort(nums.begin(), nums.end());

        for(int i = 0; i <= (int)(nums.size() - 4); ++i) 
        {

            for(int j = i + 1; j <= (int)(nums.size() - 3); ++j)
            {

                int left = j + 1, right = nums.size() - 1;
                while(left < right) 
                {
                    int sum = nums[i] + nums[j] + nums[left] + nums[right];
                    if (sum < target)
                    	++left;
                    else if (sum > target)
                    	--right;
                    else
                    {
                    	vector<int> temp = {nums[i], nums[j], nums[left], nums[right]};
                        ret.insert(temp);
                        ++left; 
                        --right;
                    } 
                }
            }
        }
        return vector<vector<int>>(ret.begin(), ret.end());
        
    }
};

方法2：手动进行去重复处理

代码

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &nums, int target)
    {

    	vector<vector<int>> ret;
        sort(nums.begin(), nums.end());
        //i直接遍历枚举四元组的第一个数下标
        for(int i = 0; i <= int(nums.size() - 4); ++i) 
        {
            
           if(i > 0 && nums[i] == nums[i - 1]) 
        		continue;
            //j直接遍历枚举四元组的第二个数下标
            for(int j = i + 1; j <= int(nums.size() - 3); ++j)
            {
                
                if(j > i + 1 && nums[j] == nums[j - 1]) 
                	continue;
              
                int left = j + 1, right = nums.size() - 1;

                while(left < right) 
                {
                    int sum = nums[i] + nums[j] + nums[left] + nums[right];
                    if(sum < target)
                    	++left;
                    else if (sum > target)
                    	--right;
                    else
                    {
                    	vector<int> temp = {nums[i], nums[j], nums[left], nums[right]};
                        ret.push_back(temp);
                        ++left;//下一个候选位置 
                        --right;//下一个候选位置
                        //对候选位置合法性进行检查
                        while(left<right && nums[left]==nums[left-1]) ++left;
                        while(left<right && nums[right]==nums[right+1]) --right;

                    } 
                }
            }
        }

        return ret;
        
    }
};