LeetCode——4Sum

1. Question

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]
]

2. Solution

这个问题的解法和3sum一样，只不过现在是4个数求和了，先遍历所有两个数字的组合，然后在去找另外两个数，时间复杂度为O(n^3)。

3. Code

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        if (nums.size() <= 3)
            return vector<vector<int>>();
        sort(nums.begin(), nums.end());
        
        vector<vector<int>> res;
        for (int i = 0; i < nums.size() - 3; i++) {
            for (int j = i + 1; j < nums.size() - 2; j++) {
                int start = j + 1;
                int end = nums.size() - 1;
                int tmp_target = target - (nums[i] + nums[j]);
                while (start < end) {
                    int tmp = nums[start] + nums[end];
                    if (tmp > tmp_target)
                        end--;
                    else if (tmp < tmp_target)
                        start++;
                    else {
                        vector<int> result;
                        result.push_back(nums[i]);
                        result.push_back(nums[j]);
                        result.push_back(nums[start]);
                        result.push_back(nums[end]);
                        res.push_back(result);
                        
                        // 去掉start开始重复的
                        while (start < end && nums[start] == result[2])
                            start++;
                        // 去掉end开始重复的
                        while (end > start && nums[end] == result[3])
                            end--;
                    }
                }
                // 去掉重复的第二个数
                while (j + 1 < nums.size() - 2 && nums[j + 1] == nums[j])
                    j++;
            }
            // 去掉重复的第一个数
            while (i + 1 < nums.size() - 3 && nums[i + 1] == nums[i])
                i++;
        }
        return res;
    }
};