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leetcode-4Sum

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Leetcode-4Sum

此题不难，但编码起来，许多细节需要注意．wrong了好多次，发现有两处画蛇添足，良久才恍然大悟．
有时候，思维逻辑正确，也不能保证编码不出小差错．

Problem:

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, a ≤ b ≤ c ≤ d)

The solution set must not contain duplicate quadruplets.

For example, given array S = {1 0 -1 0 -2 2}, and target = 0.

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A solution set is:
(-1,  0, 0, 1)
(-2, -1, 1, 2)
(-2,  0, 0, 2)
Code:
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class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target)
    {
        int len = num.size();
        sort(num.begin(),num.end());
        int i,j,k,l;
        vector<vector<int>>result;
        vector<int> vec;
        int sum;
        for(i=0;i<len-3;i++)
        {
            if(i>0&&num[i]==num[i-1])
            {
              //  i++;
                continue;
            }
            for(j=i+1;j<len-2;j++)
            {
                if(j>i+1&&num[j]==num[j-1])
                {
                   // j++;
                    continue;
                }
                l = len-1;
                k = j+1;
                while(l>k)
                {
                    if(l<len-1&&num[l]==num[l+1])
                    {
                        l--;
                        continue;
                    }
                    if(k>j+1&&num[k]==num[k-1])
                    {
                        k++;
                        continue;
                    }
                    sum = num[i]+num[j]+num[k]+num[l];
                    if(sum>target)
                    {
                        l--;
                        continue;
                    }
                    if(sum<target)
                    {
                        k++;
                        continue;
                    }
                    vec.push_back(num[i]);
                    vec.push_back(num[j]);
                    vec.push_back(num[k]);
                    vec.push_back(num[l]);
                    result.push_back(vec);
                    vec.clear();
                    k++;
                }
            }
        }
        return result;
    }
};

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