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;    }};