<p>Given an array <code>nums</code> of <code>n</code> integers, return <em>an array of all the <strong>unique</strong> quadruplets</em> <code>[nums[a], nums[b], nums[c], nums[d]]</code> such that:</p>

<ul>
	<li><code>0 &lt;= a, b, c, d&nbsp;&lt; n</code></li>
	<li><code>a</code>, <code>b</code>, <code>c</code>, and <code>d</code> are <strong>distinct</strong>.</li>
	<li><code>nums[a] + nums[b] + nums[c] + nums[d] == target</code></li>
</ul>

<p>You may return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,0,-1,0,-2,2], target = 0
<strong>Output:</strong> [[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,2,2,2,2], target = 8
<strong>Output:</strong> [[2,2,2,2]]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 200</code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= target &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        long long t=target;
        int n=nums.size();vector<vector<int>>re;
        set<vector<int>>p;
        sort(nums.begin(),nums.end());
        for(int i=0;i<n;i++){
            if((nums[i] > target) &&(target >= 0)){break;return re;}
            for(int j=i+1;j<n;j++){
                int k1=j+1,k2=n-1;
                //cout<<i<<j<<k1<<k2;return re;
                //if(k2==i||k2==j){continue;}
                while((k1<n)&&(k1<k2)){
                    long long s=nums[i]+nums[j];
                    if(s+nums[k1]>=INT_MAX){break;return re;}else{s+=nums[k1];}
                    if(s+nums[k2]>=INT_MAX){break;return re;}else{s+=nums[k2];}
                    if(s==t){
                        vector<int>r={nums[i],nums[j],nums[k1],nums[k2]};
                        p.emplace(r);
                        k1+=1;k2-=1;
                    }
                    else if(s<t){
                        k1+=1;
                    }
                    else{
                        k2-=1;
                    }
                }
            }
        }
        for(auto x:p){re.push_back(x);}
        return re;
    }
};
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