10：4 sum

注：本题的解法思想及参考的代码来自于https://github.com/soulmachine/leetcode#leetcode题解

题目: 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)

解析1：先排序，然后左右夹逼，时间复杂度O(n^3)，空间复杂度O(1)

代码如下：

class Solution {public:        vector<vector<int>> fourSum(vector<int>& nums, const int target) {                vector<vector<int>> result;                if (nums.size() < 4) return result;                sort(nums.begin(), nums.end());                for (auto i = nums.begin(); i != nums.end() - 3; ++i) {                        if (i > nums.begin() && *i == *(i - 1)) continue;                        for (auto j = i + 1; j != nums.end() - 2; ++j) {                                if (j > i + 1 && *j == *(j - 1)) continue;                                auto k = j + 1;                                auto l = nums.end() - 1;                                while (k < l) {                                        const int sum = *i + *j + *k + *l;                                        if (sum < target)                                                ++k;                                        else if (target < sum)                                                --l;                                        else {                                                result.push_back(vector<int>{*i, *j, *k, *l});                                                ++k, --l;                                                while (k < l && *k == *(k - 1))                                                        ++k,                                                while (k < l && *l == *(l + 1))                                                        --l;                                        }                                }                        }                }                return result;        }};

解析2：可以先用一个 hashmap 先缓存两个数的和，复杂度O(n^3)，空间复杂度O(n^2)

代码如下：

class Solution {public:        vector<vector<int>> fourSum(vector<int>& nums, const int target) {                vector<vector<int>> result;                if (nums.size() < 4) return result;        //      sort(nums.begin(), nums.end());                unordered_multimap<int, pair<int, int>> cache;                for (int i = 0; i != cache.size() - 1; ++i)                        for (int j = i + 1; j != nums.size(); ++j)                                cache.insert(make_pair(nums[i] + nums[j]), make_pair(i, j));                for (auto i = cache.begin(); i != cache.end(); ++i) {                        const int gap = target - i->first;                        auto range = cache.equal_range(gap);                        for (auto j = range.first; j != range.second; ++j) {                                auto a = i->second.first;                                auto b = i->second.second;                                auto c = j->second.first;                                auto d = j->second.second;                                if (a != c && a != d && b != c && b != d) {                                        vector<int> vec{nums[a], nums[b], nums[c], nums[d]};                                        sort(vec.begin(), vec.end());                                        result.push_back(vec);                                }                        }                }                sort(result.begin(), result.end());                result.erase(unique(result.begin(), result.end()), result.end());                return result;        }};