18. 4Sum

题目描述

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: 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]
]

思路

目的数组S中找出四个元素使之和为target，与3sum类似，只是外面变为了二重循环。

注意去重

代码

class Solution {
public:
    vector<vector <int> > fourSum(vector<int>& nums, int target) {
        int i, j;
        vector<vector <int>> res;
        if (nums.size() < 4)
            return res;
        sort(nums.begin(), nums.end());
        if (nums.size() == 4)
        {
            if (nums[0] + nums[1] + nums[2] + nums[3] == target)
            {
                vector<int> ini;
                ini.push_back(nums[0]);
                ini.push_back(nums[1]);
                ini.push_back(nums[2]);
                ini.push_back(nums[3]);
                res.push_back(ini);
            }
            else
            {
            }
            return res;
        }
        for (i = 0; i < nums.size() - 3; i++)
        {
            for (j = i + 1; j < nums.size() - 2; j++)
            {
                int k = j + 1;
                int l = nums.size() - 1;
                while (k < l)
                {
                    if (nums[i] + nums[j] + nums[k] + nums[l] > target)
                    {
                        l--;
                    }
                    else if (nums[i] + nums[j] + nums[k] + nums[l] < target)
                    {
                        k++;
                    }
                    else if (nums[i] + nums[j] + nums[k] + nums[l] == target)
                    {
                        vector<int> ini;
                        ini.push_back(nums[i]);
                        ini.push_back(nums[j]);
                        ini.push_back(nums[k]);
                        ini.push_back(nums[l]);
                        res.push_back(ini);
                        k++;
                        l--;
                        while (nums[k - 1] == nums[k])
                            k++;
                        while (nums[l] == nums[l + 1])
                            l--;
                    }
                }
                while (nums[j] == nums[j + 1])
                    j++;
            }
            while (nums[i] == nums[i + 1])
                i++;
        }
        return res;
    }
};