4Sum

一. 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: 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]
]

Difficulty:Medium

TIME：15MIN

解法

經歷了3Sum的洗禮後，這道題也變得異常簡單，當然除了複雜度稍微提高外，這道題和3Sum的解法幾乎是如出一轍。

vector<vector<int>> fourSum(vector<int>& nums, int target) {
    vector<vector<int>> v;
    if (nums.size() < 4)
        return v;
    sort(nums.begin(), nums.end());
    for (int i = 0; i < nums.size() - 3; i++) {
        for (int j = i + 1; j < nums.size() - 2; j++) {
            int left = j + 1;
            int right = nums.size() - 1;
            while (left < right) {
                if (nums[i] + nums[j] + nums[left] + nums[right] < target)
                    left++;
                else if (nums[i] + nums[j] + nums[left] + nums[right] > target)
                    right--;
                else {
                    vector<int> tmp{ nums[i],nums[j],nums[left],nums[right] };
                    v.push_back(tmp);
                    while (left + 1 < right && nums[left + 1] == nums[left])
                        left++;
                    while (left + 1 < right && nums[right - 1] == nums[right])
                        right--;
                    left++;
                    right--;
                }
            }
            while (nums[j + 1] == nums[j])
                j++;
        }
        while (nums[i + 1] == nums[i])
            i++;
    }
    return v;
}

代碼的時間複雜度爲O(n3) 。