4Sum

最新推荐文章于 2021-08-20 00:33:10 发布

yuluo_csdn

最新推荐文章于 2021-08-20 00:33:10 发布

阅读量383

点赞数

分类专栏： ACM 文章标签： c

本文链接：https://blog.youkuaiyun.com/yuluo_csdn/article/details/46952561

版权

ACM 专栏收录该内容

13 篇文章

订阅专栏

最近较忙，没怎么写LeetCode以后尽量保证每天一道题目以上。
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)

最近忙别的事情，看不懂代码的直接邮箱联系yuluoyzw@gmail.com.

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int>> rsvv;
        if (nums.size()<4)
            return rsvv;
        vector<int> temp(4, 0);
        sort(nums.begin(), nums.end());
        int size = nums.size();
        for (int i = 0; i<size - 3; i++)
        for (int j = i + 1; j<size- 2; j++)
        {
            int m = j + 1, n = size - 1;
            while (m<n)
            {
                int sum = nums[i] + nums[j] + nums[m] + nums[n];
                if (sum<target)
                    m++;
                else if (sum>target)
                    n--;
                else{
                    temp[0] = nums[i];
                    temp[1] = nums[j];
                    temp[2] = nums[m];
                    temp[3] = nums[n];
                    //if (!(rsvv.size()>0 && rsvv[rsvv.size() - 1][0] == temp[0] && rsvv[rsvv.size() - 1][1] == temp[1]
                        //&& rsvv[rsvv.size() - 1][2] == temp[2] && rsvv[rsvv.size() - 1][3] == temp[3]))
                        //rsvv.push_back(temp);
                    if (!((j > i + 1 && nums[j] == nums[j - 1]) || (m > j + 1 && nums[m] == nums[m - 1]) || i > 0 && nums[i] == nums[i - 1]))
                    {
                        rsvv.push_back(temp);
                    }
                    m++;

                }
            }
        }
        return rsvv;
    }
};