問題描述

給定一個數組arr，以及一個數字target。求陣列中和為target的三個數。要求給出所有解。陣列中的每個數在每組解中只能出現一次。

解法

方法一：如果用暴力方法，時間複雜度是o(n3)

方法二：利用C++11中的unordered_multimap，本題複雜度可降至O(n2)

#include<iostream>
#include<vector>
#include<unordered_map>
#include <set>
#include <algorithm>
using namespace std
 
;
class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num, int sum) {
        unordered_multimap<int, int> hashmap;
        vector<vector<int> > result;
        for(int i=num.size()-1; i>=0; i--)
        {
            hashmap.insert( {num.at(i), num.size()-i-1
 
} );
        }
        /*
        for(auto iter1=hashmap.begin(); iter1!=hashmap.end(); iter1++)
        {
            cout<<(*iter1).first<<" ";
            cout<<(*iter1).second<<endl;
        }
        cout<<endl;
        */
        for(auto iter1=hashmap.begin(); iter1!=hashmap.end(); iter1++)
        {
            int
 
 target = sum -(*iter1).first;
            auto iter = iter1;
            iter++;
            if (iter == hashmap.end() )
                continue;
            for(auto iter2=iter; iter2!=hashmap.end(); iter2++)
            {
                auto iterres = hashmap.equal_range( target - (*iter2).first );
                if (iterres.first == iterres.second)//下界pair和上界pair相等，沒找到
                {
                    continue;
                }
                for(auto get=iterres.first; get!=iterres.second; get++)
                {
                    //get遍歷所有pair(一對mapiterator),(*get)取出一個pair,(*get).second表示第二個map變數
                    //if (get == hashmap.end() || (*iter1).second >= (*iter2).second || (*iter2).second >= (*get).second )
                    if (get == hashmap.end() || (*iter1).second == (*iter2).second || (*iter2).second == (*get).second || (*iter1).second == (*get).second ) //後面三個都不能遺漏！
                    {
                        continue;
                    }
                    int res = (*get).first;
                    int arr[] = { (*iter2).first, (*iter1).first, res};
                    sort(arr, arr+3);
                    vector<int> trip(arr, arr+3);
                    result.push_back(trip);
                }
            }
        }
        set< vector<int> > tmpres;
        for(int i=0; i<result.size(); i++)
        {
            tmpres.insert( result.at(i) );
        }
        result.clear();
        for(auto iter=tmpres.begin(); iter!=tmpres.end(); iter++)
        {
            result.push_back( *iter );
        }
        return result;
    }
};
int main()
{
    Solution sol;
    //int arr[] = {-1, 0, 1, 2, -1, -4};
    int arr[] = {7,5,-8,-6,-13,7,10,1,1,-4,-14,0,-1,-10,1,-13,-4,6,-11,8,-6,0,0,-5,0,11,-9,8,2,-6,4,-14,6,4,-5,0,-12,12,-13,5,-6,10,-10,0,7,-2,-5,-12,12,-9,12,-9,6,-11,1,14,8,-1,7,-13,8,-11,-11,0,0,-1,-15,3,-11,9,-7,-10,4,-2,5,-4,12,7,-8,9,14,-11,7,5,-15,-15,-4,0,0,-11,3,-15,-15,7,0,0,13,-7,-12,9,9,-3,14,-1,2,5,2,-9,-3,1,7,-12,-3,-1,1,-2,0,12,5,7,8,-7,7,8,7,-15};
    vector<int> num(arr, arr+200);
    vector<vector<int> > res = sol.threeSum(num, 10);
    for(int i=0; i<res.size(); i++)
    {
        for(int j=0; j<res.at(i).size(); j++)
        {
            cout<<res.at(i).at(j)<<" ";
        }
        cout<<endl;
    }
    return 0;
}

以上解法（稍加修改），提交給leetcode（第15題）後仍然報錯（Time Limit Exceeded）。難道有更低複雜度的解法嗎？
希望有高手能指點一二。

方法三：先排序，然後二分查詢，複雜度可降至O(nlogn)

 vector<vector<int> > threeSum(vector<int> &num) 
 {
        vector<vector<int> > res;
        if(num.empty())
            return res;
        sort(num.begin(), num.end()); //nlogn

        for(int i=0; i+2<num.size(); i++) //n
        {
            if(i>0 && num[i] == num[i-1])
                continue;
            int left(i+1), right(num.size()-1);
            while(left < right)//logn
            {
                if(left>i+1 && num[left] == num[left-1])
                {
                    left++;
                    continue;
                }
                if(right < num.size()-1 && num[right] == num[right+1])
                {
                    right--;
                    continue;
                }
                int sum = num[i] + num[left] + num[right];
                if(sum < 0) left++;
                else if(sum > 0) right--;
                else
                {
                    res.push_back( vector<int> {num[i], num[left], num[right]} );
                    right--;
                    left++;
                }
            }
        }
        return res;
    }

同理，如果需要查詢陣列中和為某給定值的四元組，用法三解決，時間複雜度為O(n2logn)