LeetCode-18.四数之和

问题： 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)

方法1

类似于最接近目标的三个数值加和，先对数组进行排序，然后在内层进行左右夹逼。
时间复杂度为

#include <iostream>
#include <vector>
#include <algorithm>
#include <unordered_map>
using namespace std;

//时间复杂度为O(n^3)，空间复杂度为O(1)
vector<vector<int>> fourSum_1(vector<int>& nums, int target) {
    vector<vector<int>> result;
    if (nums.size() < 4)
        return result;

    sort(nums.begin(), nums.end());

    for (int i = 0; i < nums.size() - 3; i++) {
        if (i > 0 && nums[i] == nums[i - 1])
            continue;
        for (int j = i + 1; j < nums.size(); j++) {
            if (j > i + 1 && nums[j] == nums[j - 1])
                continue;
            int k = j + 1;
            int l = nums.size() - 1;
            while (k < l) {
                const int sum = nums[i] + nums[j] + nums[k] + nums[l];
                if (sum < target) {
                    k++;
                    while (nums[k] == nums[k - 1] && k < l)
                        k++;
                }
                else if (sum > target) {
                    l--;
                    while (nums[l] == nums[l + 1] && k < l)
                        l--;
                }
                else {
                    result.push_back({ nums[i],nums[j],nums[k],nums[l] });
                    k++;
                    l--;
                    while (nums[k] == nums[k - 1] && k < l)
                        k++;
                    while (nums[l] == nums[l + 1] && k < l)
                        l--;
                }
            }
        }
    }
    return result;
}

方法2

使用HashMap缓存两个数的和，然后再将这些和进行配对，最后剔除重复的组合。

#include <iostream>
#include <vector>
#include <algorithm>
#include <unordered_map>
using namespace std;

//时间复杂度：平均O(n^2)，最坏O(n^4)，空间复杂度O(n^2)
vector<vector<int>> fourSum_2(vector<int>& nums, int target) {
    vector<vector<int>> result;
    if (nums.size() < 4)
        return result;
    sort(nums.begin(), nums.end());

    unordered_map<int, vector<pair<int, int>>> cache;
    for (size_t a = 0; a < nums.size(); a++) {
        for (size_t b = a + 1; b < nums.size(); b++) {
            //cache存储形式：（sum,[<value,value>,<value,value>]）
            cache[nums[a] + nums[b]].push_back(pair<int, int>(a, b));
        }
    }

    for (int c = 0; c < nums.size(); c++) {
        for (size_t d = c + 1; d < nums.size(); d++) {
            const int key = target - nums[c] - nums[d];
            if (cache.find(key) == cache.end())
                continue;
            //获取vector，其中pair元素能和当前pair元素和等于target
            const auto &vec = cache[key];
            for (size_t k = 0; k < vec.size(); k++) {
                //<nums[c],num[d]>和vec[k]进行比较
                //去掉重复
                if (c <= vec[k].second)
                    continue;
                result.push_back({ nums[vec[k].first],nums[vec[k].second],nums[c],nums[d] });
            }
        }
    }
    sort(result.begin(), result.end());
    result.erase(unique(result.begin(), result.end()), result.end());
    return result;
}