18. 4Sum**

18. 4Sum**

​​https://leetcode.com/problems/4sum/description/​​

题目描述

Given an array ​​nums​​​ of ​​n​​​ integers and an integer ​​target​​​, are there elements ​​a, b, c​​​ and ​​d​​​ in ​​nums​​​ 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]
]

给定一个整数数组和给定的 ​​target​​​, 是否存在 ​​a, b, c, d​​​ 4个整数使得它们的和满足 ​​a+b+c+d = target​​​? 返回所有不重复的 ​​(a, b, c, d)​​ 四元组.

解题思路

首先固定 a, 然后再使用 ​​3Sum​​ 的方法求解.

C++ 实现 1

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        if (nums.size() < 4)
            return {};

        std::sort(nums.begin(), nums.end());
        vector<vector<int>> res;
        vector<int> path;
        for (int i = 0; i < nums.size() - 3; ++i) {
          for (int j = i + 1; j < nums.size() - 2; ++j) {
            int lo = j + 1, hi = nums.size() - 1;
            while (lo < hi) {
              int sum = nums[i] + nums[j] + nums[lo] + nums[hi];
              if (sum == target) {
                path = {nums[i], nums[j], nums[lo], nums[hi]};
                res.push_back(path);
                while (lo < hi && nums[lo] == nums[lo + 1]) lo ++;
                while (lo < hi && nums[hi] == nums[hi - 1]) hi --;
                lo ++;
                hi --;
              }
              else if (sum < target)
                lo ++;
              else
                hi --;
            }
            while (j + 1 < nums.size() - 2 && nums[j + 1] == nums[j]) ++j;
          }
          while (i + 1 < nums.size() - 3 && nums[i + 1] == nums[i]) ++i;
        }
        return res;
    }
};

C++ 实现 2

注意不要忘记先给数组排序, 此外还需要过滤重复元素.

class Solution {
private:
    vector<vector<int>> threeSum(vector<int> &nums, int i, int j, int target) {
        if (nums.size() < 3 || j - i < 2) return {};
        vector<vector<int>> res;
        for (int k = i; k < j - 1; ++ k) {
            if (k == i || (k > i && nums[k] != nums[k - 1])) {
                int lo = k + 1, hi = j;
                while (lo < hi) {
                    int sum = nums[lo] + nums[hi] + nums[k];
                    if (sum == target) {
                        res.push_back({nums[k], nums[lo], nums[hi]});
                        while (lo + 1 < hi && nums[lo + 1] == nums[lo]) ++ lo;
                        while (hi - 1 > lo && nums[hi - 1] == nums[hi]) -- hi;
                        ++ lo;
                        -- hi;
                    } else if (sum < target) ++ lo;
                    else -- hi;
                }
            }
        }
        return res;
    }
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        if (nums.size() < 4) return {};
        std::sort(nums.begin(), nums.end());
        vector<vector<int>> res;
        for (int i = 0; i < nums.size() - 3; ++ i) {
            if (i == 0 || (i > 0 && nums[i] != nums[i - 1])) {
                int aim = target - nums[i];
                auto pairs = threeSum(nums, i + 1, nums.size() - 1, aim);
                if (!pairs.empty())
                    for (auto &pair : pairs)
                        res.push_back({nums[i], pair[0], pair[1], pair[2]});
            }
        }
        return res;
    }
};

C++ 实现 3

过滤重复元素的方法和上一节不同.

class Solution {
private:
    vector<vector<int>> twoSum(vector<int> &nums, int start, int target) {
        vector<vector<int>> res;
        int i = start, j = nums.size() - 1;
        while (i < j) {
            if (nums[i] + nums[j] == target) {
                res.push_back({nums[i], nums[j]});
                while (i < j && nums[i + 1] == nums[i]) i ++;
                while (i < j && nums[j - 1] == nums[j]) j --;
                i ++;
                j --;
            }
            else if (nums[i] + nums[j] < target)
                ++ i;
            else
                -- j;
        }
        return res;
    }
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        if (nums.size() < 4)
            return {};

        vector<vector<int>> res;
        std::sort(nums.begin(), nums.end());
        for (int i = 0; i < nums.size() - 3; ++i) {
            for (int j = i + 1; j < nums.size() - 2; ++j) {
                auto temp = twoSum(nums, j + 1, target - nums[i] - nums[j]);
                if (!temp.empty())
                    for (auto &v : temp)
                        res.push_back({nums[i], nums[j], v[0], v[1]});

                while (j + 1 < nums.size() - 2 && nums[j + 1] == nums[j]) ++j;
            }
            while (i + 1 < nums.size() - 3 && nums[i + 1] == nums[i]) ++i;
        }
        return res;
    }
};