4Sum

最新推荐文章于 2018-10-30 15:51:24 发布

Pango_lulu

最新推荐文章于 2018-10-30 15:51:24 发布

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分类专栏： Leetcode 文章标签： array two pointer

本文链接：https://blog.youkuaiyun.com/pango_cs/article/details/41857079

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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.
    A solution set is:
    (-1, 0, 0, 1)
    (-2, -1, 1, 2)
    (-2, 0, 0, 2)

vector<vector<int> > fourSum(vector<int> &num, int target)
{
    vector<vector<int> > ans;
    ans.clear();
    sort(num.begin(), num.end());

    for(int i = 0; i < (int)num.size() - 3; i++)
    {
        if(i >= 1 && num[i] == num[i-1])
            continue;

        for(int j = i + 1; j < (int)num.size() - 2; j++)
        {
            if(j > i + 1 && num[j] == num[j-1])
                continue;

            int left = j + 1;
            int right = num.size() - 1;
            while(left < right)
            {
                if(left > j + 1 && num[left] == num[left - 1])
                {
                    left++;
                    continue;
                }
                if(right < num.size()-1 && num[right] == num[right + 1])
                {
                    right--;
                    continue;
                }
                if(num[i] + num[j] + num[left] + num[right] < target)
                    left++;
                else if(num[i] + num[j] + num[left] + num[right] > target)
                    right--;
                else
                {
                    vector<int> tmp;
                    tmp.clear();
                    tmp.push_back(num[i]);
                    tmp.push_back(num[j]);
                    tmp.push_back(num[left]);
                    tmp.push_back(num[right]);
                    ans.push_back(tmp);
                    left++;
                    right--;
                }
            }
        }
    }
    return ans;
}