• 4Sum_leetCode


    Given an array S of 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, abcd)
    • 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)
    

    点击打开原题链接


    N SUM和都是一个德性,代码例如以下:

    class Solution 
    {
    private:
        vector<vector<int> > ret;
    public:
        vector<vector<int> > fourSum(vector<int> &num, int target) 
        {
            sort(num.begin(), num.end());
            
            ret.clear();
            
            for(int i = 0; i < num.size(); i++)
            {
                if (i > 0 && num[i] == num[i-1])
                    continue;
                    
                for(int j = i + 1; j < num.size(); j++)
                {
                    if (j > i + 1 && num[j] == num[j-1])
                        continue;
                        
                    int k = j + 1;
                    int t = num.size() - 1;
                    
                    while(k < t)
                    {
                        if (k > j + 1 && num[k] == num[k-1])
                        {
                            k++;
                            continue;
                        }
                        
                        if (t < num.size() - 1 && num[t] == num[t+1])
                        {
                            t--;
                            continue;
                        }
                        
                        int sum = num[i] + num[j] + num[k] + num[t];
                        
                        if (sum == target)
                        {
                            vector<int> a;
                            a.push_back(num[i]);
                            a.push_back(num[j]);
                            a.push_back(num[k]);
                            a.push_back(num[t]);
                            ret.push_back(a);
                            k++;
                            t--;
                        }
                        else if (sum < target)
                            k++;
                        else
                            t--;                        
                    }
                }
            }
            
            return ret;
        }
    };



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  • 原文地址:https://www.cnblogs.com/liguangsunls/p/6777329.html
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