code第一部分数组：第九题 四个元素之和为给定目标

code第一部分数组：第九题  四个元素之和为给定目标

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)

解决方案1 先排序，然后二分查找，复杂度 O(n3 log n)

vector<vector<int>> fourSum(vector<int>& num, int target)
{
    vector<vector<int>> result;
    if (num.size() < 4)
        return result;
    sort(num.begin(), num.end());
    auto last = num.end();
    for (auto a = num.begin(); a < prev(last, 3);a = upper_bound(a, prev(last, 3), *a))
    {
        for (auto b = next(a); b < prev(last, 2);b = upper_bound(b, prev(last, 2), *b))
        {
            for (auto c = next(b); c < prev(last);c = upper_bound(c, prev(last), *c))
            {
                const int d = target - *a - *b - *c;
                if (binary_search(next(c), last, d))
                result.push_back(vector<int> { *a, *b, *c, d });
            }
        }
    }
    return result;
}

方法2 使用map做缓存，先缓存两个数的和，时间复杂度 O(n^3)，空间复杂度 O(n^2)

这个方法不会！