[leetcode 18] 4Sum

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本文介绍了一种解决四数之和问题的有效算法。通过先排序再使用双指针技巧，并借助哈希表来减少时间复杂度，将问题转换为两数之和。确保输出的四元组递增有序且不包含重复。

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;
    }
};