如何在 3Sum 问题中返回所有可能的三元组？

Question

我正在做 3Sum 问题：https://leetcode.com/explore/interview/card/top-interview-questions-medium/103/array-and-strings/776/

简单问题：给定一个整数数组 nums，返回所有三元组 [nums[i], nums[j], nums[k]]，使得 i != j、i != k、j != k 和 nums[i] + nums[j] + nums[k] == 0。 请注意，解集不得包含重复的三元组。

我的问题：我有一个解决方案，它确实返回了一些但不是所有可能的三元组，我不明白我哪里出错了。其次，我的算法是O(N^2 log N)，欢迎提出改进建议。

Input:
[-1,0,1,2,-1,-4,-2,-3,3,0,4]
Output:
[[-3,-1,4],[-3,0,3],[-4,1,3],[-2,0,2],[-4,0,4]]
Expected:
[[-4,0,4],[-4,1,3],[-3,-1,4],[-3,0,3],[-3,1,2],[-2,-1,3],[-2,0,2],[-1,-1,2],[-1,0,1]]

算法：我已经将我的算法与 cmets 一起包含在代码中，但这里是要点 - 对于每对数字，我将它们的 sum 存储为 key 以及给我的数字的索引 @ 987654330@ 作为value。然后在一个循环中，我遍历每个元素并检查target 值与该数字之间的差异是否以key 的形式出现在map 中。如果是，并且所有索引都不相等，我将其添加到最终返回的 vector。

代码：

#include <iostream>
#include <vector>
#include <unordered_map>

using namespace std;

vector<vector<int>> threeSum(vector<int> &nums) {
    /*
     * If less than 3 elements are passed return empty vector
     */
    if (nums.size() < 3)
        return {};

    int target = 0; // all three numbers sum up to 0 as given
    vector<vector<int>> outer; // this is returned by the function
    unordered_map<int, vector<int>> umap; // unordered_map for keeping sum and indices

    /*
     * Calculate sum of each pair of numbers
     * Store the sum as key and the pair as value
     * i != j is guaranteed
     */
    for (int i = 0; i < nums.size(); i++)
        for (int j = i + 1; j < nums.size(); j++)
            umap[nums[i] + nums[j]] = {i, j};

    /*
     * Go through each element and calculate the difference
     * between target and that element, this gives the sum
     * of the other two elements.
     * Look for the sum in unordered_map
     * If it is present check if all three indices are not equal to each other
     */
    for (int i = 0; i < nums.size(); i++) {
        vector<int> inner;
        int diff = target - nums[i];
        auto it = umap.find(diff);
        inner = umap[diff];
        if (it != umap.end() && i != inner[0] && i != inner[1]) {
            inner.push_back(i);
            vector<int> tmp;
            for (auto &j: inner)
                tmp.push_back(nums[j]); // push actual numbers instead of indices
            sort(tmp.begin(), tmp.end()); // sort the inner three elements
            if (find(outer.begin(), outer.end(), tmp) == outer.end()) // for removing duplicates
                outer.push_back(tmp);
        }
    }
    return outer;
}

int main() {
    vector<int> v{-1, 0, 1, 2, -1, -4, -2, -3, 3, 0, 4};
    vector<vector<int>> ret = threeSum(v);
    for (auto &i: ret) {
        for (auto j: i)
            cout << j << " ";
        cout << endl;
    }
}

Answer 1

您的错误通过输入 [-5, 1, 2, 3, 4] 进行演示。

问题是5 == 1+4 == 2+3。您的 map 只会将其中一对存储为该键的值，因此它不会记录这两个对都可以导致该键的事实。

Answer 2

既然你要求改进这个算法怎么样：

创建一个std::unordered_map<int, int> 将每个输入映射到出现次数。那么：

for (const auto & it = map.begin(); it != map.end(); ++it) {
    for (const auto & it2 = it; it2 != map.end(); ++it2) {
        auto it3 = map.find(-*it - *it2);
        if (it3 != map.end()) {
            it ((*it > *it3) || (*it2 > *it3)) continue;
            if (it == it2) {
                if (it == it3) {
                   if (it.second > 2) cout << *it << ", " *it2 << ", " << *it3 << std::endl;
                } else {
                   if (it.second > 1) cout << *it << ", " *it2 << ", " << *it3 << std::endl;
                }
            } else if (it2 == it3) {
                if (it2.second > 1) cout << *it << ", " *it2 << ", " << *it3 << std::endl;
            } else {
                cout << *it << ", " *it2 << ", " << *it3 << std::endl;
            }
        }             
    }
}

应该是O(n^2)，没有重复。

Answer 3

我确实解决了这个问题，但它效率低下，并且在一些测试用例上超出了时间限制。

#include <iostream>
#include <vector>
#include <unordered_map>
#include <set>

using namespace std;

vector<vector<int>> threeSum(vector<int> &nums) {
    /*
     * If less than 3 elements are passed return empty vector
     */
    if (nums.size() < 3)
        return {};

    int target = 0; // all three numbers sum up to 0 as given
    vector<vector<int>> outer; // this is returned by the function
    unordered_multimap<int, vector<int>> umap; // unordered_map for keeping sum and indices

    /*
     * Calculate sum of each pair of numbers
     * Store the sum as key and the pair as value
     * i != j is guaranteed
     */
    for (int i = 0; i < nums.size(); i++)
        for (int j = i + 1; j < nums.size(); j++) {
            vector<int> temp;
            temp.push_back(i);
            temp.push_back(j);
            umap.emplace(make_pair(nums[i] + nums[j], temp));
        }

    /*
     * Go through each element and calculate the difference
     * between target and that element, this gives the sum
     * of the other two elements.
     * Look for the sum in unordered_map
     * If it is present check if all three indices are not equal to each other
     */
    //unordered_set<vector<int>> uset;
    for (int i = 0; i < nums.size(); i++) {
        vector<int> inner;
        int diff = target - nums[i];
        auto it = umap.find(diff);
        while (it != umap.end()) {
            inner = it->second;
            if (it != umap.end() && i != inner[0] && i != inner[1]) {
                inner.push_back(i);
                //vector<int> tmp;
                //for (auto &j: inner)
                //    tmp.push_back(nums[j]); // push actual numbers instead of indices
                //sort(tmp.begin(), tmp.end()); // sort the inner three elements
                //if (find(outer.begin(), outer.end(), tmp) == outer.end()) // for removing duplicates
                outer.push_back(inner);

            }
            umap.erase(it);
            it = umap.find(diff);
        }
    }
    for (auto &i: outer)
        for (auto &j: i)
            j = nums[j];
    //sort(i.begin(),i.end());
    set<vector<int>> uset;
    for (auto &i: outer) {
        sort(i.begin(), i.end());
        uset.insert(i);
    }
    outer.clear();
    outer.insert(outer.end(), uset.begin(), uset.end());
    return outer;
}

int main() {
    vector<int> v{-1, 0, 1, 2, -1, -4, -2, -3, 3, 0, 4};
    vector<vector<int>> ret = threeSum(v);
    for (auto &i: ret) {
        for (auto j: i)
            cout << j << " ";
        cout << endl;
    }
}