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2014
11-19

LeetCode-4Sum[数组]

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:

  • Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, abcd)
  • 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)

标签: Array Hash Table Two Pointers
分析

先排序,然后左右夹逼,复杂度 $O(n^3)$,会超时。

可以用一个hashmap先缓存两个数的和,最终复杂度$O(n^3)$。这个策略也适用于 3Sum 。

代码1

// LeetCode, 4Sum
// 先排序,然后左右夹逼,时间复杂度O(n^3),空间复杂度O(1)
class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& num, int target) {
        vector<vector<int>> result;
        if (num.size() < 4) return result;
        sort(num.begin(), num.end());

        auto last = num.end();
        for (auto a = num.begin(); a < prev(last, 3); ++a) {
            for (auto b = next(a); b < prev(last, 2); ++b) {
                auto c = next(b);
                auto d = prev(last);
                while (c < d) {
                    if (*a + *b + *c + *d < target) {
                        ++c;
                    } else if (*a + *b + *c + *d > target) {
                        --d;
                    } else {
                        result.push_back({ *a, *b, *c, *d });
                        ++c;
                        --d;
                    }
                }
            }
        }
        sort(result.begin(), result.end());
        result.erase(unique(result.begin(), result.end()), result.end());
        return result;
    }
};

代码2

// LeetCode, 4Sum
// 用一个hashmap先缓存两个数的和
// 时间复杂度,平均O(n^2),最坏O(n^4),空间复杂度O(n^2)
class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        vector<vector<int>> result;
        if (num.size() < 4) return result;
        sort(num.begin(), num.end());

        unordered_map<int, vector<pair<int, int> > > cache;
        for (size_t a = 0; a < num.size(); ++a) {
            for (size_t b = a + 1; b < num.size(); ++b) {
                cache[num[a] + num[b]].push_back(pair<int, int>(a, b));
            }
        }

        for (int c = 0; c < num.size(); ++c) {
            for (size_t d = c + 1; d < num.size(); ++d) {
                const int key = target - num[c] - num[d];
                if (cache.find(key) == cache.end()) continue;

                const auto& vec = cache[key];
                for (size_t k = 0; k < vec.size(); ++k) {
                    if (c <= vec[k].second)
                        continue; // 有重叠

                    result.push_back( { num[vec[k].first],
                            num[vec[k].second], num[c], num[d] });
                }
            }
        }
        sort(result.begin(), result.end());
        result.erase(unique(result.begin(), result.end()), result.end());
        return result;
    }
};

代码3

// LeetCode, 4Sum
// 用一个 hashmap 先缓存两个数的和
// 时间复杂度O(n^2),空间复杂度O(n^2)
// @author 龚陆安(http://weibo.com/luangong)
class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& num, int target) {
        vector<vector<int>> result;
        if (num.size() < 4) return result;
        sort(num.begin(), num.end());

        unordered_multimap<int, pair<int, int>> cache;
        for (int i = 0; i + 1 < num.size(); ++i)
            for (int j = i + 1; j < num.size(); ++j)
                cache.insert(make_pair(num[i] + num[j], make_pair(i, j)));

        for (auto i = cache.begin(); i != cache.end(); ++i) {
            int x = target - i->first;
            auto range = cache.equal_range(x);
            for (auto j = range.first; j != range.second; ++j) {
                auto a = i->second.first;
                auto b = i->second.second;
                auto c = j->second.first;
                auto d = j->second.second;
                if (a != c && a != d && b != c && b != d) {
                    vector<int> vec = { num[a], num[b], num[c], num[d] };
                    sort(vec.begin(), vec.end());
                    result.push_back(vec);
                }
            }
        }
        sort(result.begin(), result.end());
        result.erase(unique(result.begin(), result.end()), result.end());
        return result;
    }
};

代码4

// LeetCode, 4Sum
// 先排序,然后左右夹逼,时间复杂度O(n^3logn),空间复杂度O(1),会超时
// 跟方法1相比,表面上优化了,实际上更慢了,切记!
class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& num, int target) {
        vector<vector<int>> result;
        if (num.size() < 4) return result;
        sort(num.begin(), num.end());

        auto last = num.end();
        for (auto a = num.begin(); a < prev(last, 3);
                a = upper_bound(a, prev(last, 3), *a)) {
            for (auto b = next(a); b < prev(last, 2);
                    b = upper_bound(b, prev(last, 2), *b)) {
                auto c = next(b);
                auto d = prev(last);
                while (c < d) {
                    if (*a + *b + *c + *d < target) {
                        c = upper_bound(c, d, *c);
                    } else if (*a + *b + *c + *d > target) {
                        d = prev(lower_bound(c, d, *d));
                    } else {
                        result.push_back({ *a, *b, *c, *d });
                        c = upper_bound(c, d, *c);
                        d = prev(lower_bound(c, d, *d));
                    }
                }
            }
        }
        return result;
    }
};

  1. 在方法1里面:

    //遍历所有的边,计算入度
    for(int i=0; i<V; i++)
    {
    degree = 0;
    for (j = adj .begin(); j != adj .end(); ++j)
    {
    degree[*j]++;
    }
    }

    为什么每遍历一条链表,要首先将每个链表头的顶点的入度置为0呢?
    比如顶点5,若在顶点1、2、3、4的链表中出现过顶点5,那么要增加顶点5的入度,但是在遍历顶点5的链表时,又将顶点5的入度置为0了,那之前的从顶点1234到顶点5的边不是都没了吗?