[LeetCode 16] The closest sum of three numbers (medium) Violent dfs || Double pointer

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Code one (violent dfs):

It is a miracle to pass, theoretically it takes up to 2 ¹⁰⁰⁰ times

class Solution {
    
    
public:
    int ans=1<<27; //初始为很大的数

    void dfs(int index,int sum,int k,vector<int>& nums, int target)
    {
    
    
        if(k==3 && abs(sum-target)<=abs(ans-target)) //三数 且更接近target
        {
    
    
            ans=sum;
            return;
        }

        if(index>=nums.size()) return; //越界

        dfs(index+1,sum,k,nums,target); //不要当前值
        dfs(index+1,sum+nums[index],k+1,nums,target); //要当前值
    }

    int threeSumClosest(vector<int>& nums, int target) {
    
    
        dfs(0,0,0,nums,target);
        return ans;
    }
};

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Code two double pointer:

class Solution {
    
    
public:
    int threeSumClosest(vector<int>& nums, int target) {
    
    
        sort(nums.begin(),nums.end()); //排序
        int len=nums.size(),ans=1<<27;

        for(int i=0;i<len-2;i++)
        {
    
    
            int l=i+1,r=len-1; 
            while(l<r)
            {
    
    
                int sum=nums[l]+nums[r]+nums[i];
                if(abs(ans-target)>abs(sum-target)) //更接近target
                    ans=sum;
                if(ans==target) return ans; //最小差值为0直接返回
                if(sum>target) r--; //大了
                else l++; //小了
            }
        }
        return ans;
    }
};