求无序数组中四元素相加等于特定值的所有情况

/*************************************************************************************************description:给定一个无序数组和目标数target，要求返回数组中4个元素相加等于target的所有4元素组合**            要求：4个元素从小到大排列；不能返回重复的4元素组合**            如:Input: a={1 0 -1 0 -2 2},target=0**               Output: (-1,0,0,1), (-2,-1,1,2),(-2,0,0,2)***********************************************************************************************/#include<iostream>#include<vector>#include<algorithm>#include<unordered_map>using namespace std;//方法一:先排序，然后对内层左右夹逼//时间复杂度为O(n^3)vector<vector<int>> fourSum_1(vector<int> &arr, int target){    vector<vector<int>> result;    if (arr.size() < 4)        return result;    sort(arr.begin(), arr.end());    for (int i = 0; i < arr.size() - 3; i++)    {        if (i > 0 && arr[i] == arr[i - 1])            continue;        for (int j = i + 1; j < arr.size() - 2; j++)        {            if (j > i + 1 && arr[j] == arr[j - 1])                continue;            int m = j + 1;            int n = arr.size() - 1;            while (m < n)            {                if (arr[i] + arr[j] + arr[m] + arr[n] < target)                    m++;                else if (arr[i] + arr[j] + arr[m] + arr[n] > target)                    n--;                else                {                    vector<int> group;                    group.push_back(arr[i]);                    group.push_back(arr[j]);                    group.push_back(arr[m]);                    group.push_back(arr[n]);                    result.push_back(group);                    m++;                    n--;                    while (arr[m] == arr[m -1])                        m++;                }            }        }    }    return result;}//方法二：用hashmap存储两个数之和，然后再两层遍历从hashmap中找到对应的数据//时间复杂度：平均O(n^2)，最坏O(n^4)，空间复杂度：O(n^2)vector<vector<int>> fourSum_2(vector<int> &arr, int target){    vector<vector<int>> result;    if (arr.size() < 4)        return result;    sort(arr.begin(), arr.end());    unordered_map<int, vector<pair<int, int>>> cache;    for (int i = 0; i < arr.size() - 1; i++)    {        for (int j = i + 1; j < arr.size(); j++)        {            cache[arr[i] + arr[j]].push_back(pair<int, int>(i, j));        }    }    for (int i = 0; i < arr.size(); i++)    {        for (int j = i + 1; j < arr.size(); j++)        {            int data = target - arr[i] - arr[j];            if (cache.find(data) != cache.end())            {                for (int k = 0; k < cache[data].size(); k++)                {                    if (i > cache[data][k].first  && j > cache[data][k].second && i > cache[data][k].second)                    {                        vector<int> group;                        group.push_back(arr[cache[data][k].first]);                        group.push_back(arr[cache[data][k].second]);                        group.push_back(arr[i]);                        group.push_back(arr[j]);                        result.push_back(group);                    }                }            }        }    }    return result;}//方法三：用multi hashmap存储两个数之和，然后再两层遍历从hashmap中找到对应的数据//时间复杂度：O(n^2)，空间复杂度：O(n^2)vector<vector<int>> fourSum_3(vector<int> &arr, int target){    vector<vector<int>> result;    if (arr.size() < 4)        return result;    sort(arr.begin(), arr.end());    unordered_multimap<int, pair<int, int>> cache;    for (int i = 0; i < arr.size() - 1; i++)    {        for (int j = i + 1; j < arr.size(); j++)        {            cache.insert(pair<int, pair<int,int>>(arr[i] + arr[j],pair<int,int>(i,j)));        }    }    for (auto i = cache.begin(); i != cache.end(); i++)    {        int data = target - i->first;        auto range = cache.equal_range(data);        for (auto j = range.first; j != range.second; j++)        {            int a = i->second.first;            int b = i->second.second;            int c = j->second.first;            int d = j->second.first;            if (c > b)            {                vector<int> group;                group.push_back(arr[a]);                group.push_back(arr[b]);                group.push_back(arr[c]);                group.push_back(arr[d]);                result.push_back(group);            }        }    }    result.erase(unique(result.begin(), result.end()), result.end());    sort(result.begin(), result.end());    return result;}