【Leetcode】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)

class Solution {
public:
    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) {
            if (a != num.begin() && *a == *prev(a)) continue;
            for (auto b = next(a); b < prev(last, 2); ++b) {
                if (b != next(a) && *b == *prev(b)) continue;
                auto c = next(b);
                auto d = prev(last);
                while (c < d) {
                    int sum = *a + *b + *c + *d;
                    if (sum < target) {
                        ++c;
                    } else if (sum > target) {
                        --d;
                    } else {
                        result.push_back({*a, *b, *c, *d});
                        ++c, --d;
                    }
                }
            }
        }
        result.erase(unique(result.begin(), result.end()), result.end());
        return result;
    }
};

与前面的题类似，先排序再夹逼，但是需要循环两次。O(n³).

其它的方法：hash_map缓存两两元素的和。

C++11的写法好别扭...