算法分析与设计课程（17）：【leetcode】4Sum

Description:

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: 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]
]

算法分析：

可以采取下面的步骤

1. 对数组排序
2. 确定四元数中的两个
3. 遍历剩余数组确定两外两个

算法时间复杂度为O(n^3)。

代码如下：

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &nums, int target) {
        set<vector<int> > res;
        sort(nums.begin(), nums.end());
        for (int i = 0; i < int(nums.size() - 3); ++i) {
            for (int j = i + 1; j < int(nums.size() - 2); ++j) {
                int left = j + 1, right = nums.size() - 1;
                while (left < right) {
                    int sum = nums[i] + nums[j] + nums[left] + nums[right];
                    if (sum == target) {
                        vector<int> out;
                        out.push_back(nums[i]);
                        out.push_back(nums[j]);
                        out.push_back(nums[left]);
                        out.push_back(nums[right]);
                        res.insert(out);
                        ++left; --right;
                    } else if (sum < target) ++left;
                    else --right;
                }
            }
        }
        return vector<vector<int> > (res.begin(), res.end());
    }
};