[leetcode 18] 4Sum

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)

思路：先排序，在用夹逼。为了降低时间复杂度，用哈希表保存两个数的和作为键，其下标为值，转化为2Sum。

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