4Sum--LeetCode

1.题目

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: 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.题意

查找数组中和为０的４个数
解决方案集不能含有重复项

3.分析

1） 传统k-sum问题，先排序，再左右夹逼
时间复杂度O(n^3)，空间复杂度O(1)
2）借助unordered_multimap（允许重复key）缓存两数之和
时间复杂度O(n^2)，空间复杂度O(n^2)
注意内层循环是for(int j = i + 1; j < nums.size(); ++j)而非for(int j = 0; j < nums.size(); ++j)
a = i->second.first;而非a = i->first.first;
if(a != c && a != d && b != c && b != d)而非if(nums[a] + nums[b] + nums[c] + nums[d] == target)

4.代码

1）

class Solution {public:    vector<vector<int>> fourSum(vector<int>& nums, int target) {        vector<vector<int>> result;        if(nums.size() < 4)            return result;        sort(nums.begin(), nums.end());        int len = nums.size();        for(int a = 0; a < len - 3; ++a)        {            for(int b = a + 1; b < len - 2; ++b)            {                int c = b + 1;                int d = len - 1;                while(c < d)                {                    if(nums[a] + nums[b] + nums[c] + nums[d] < target)                        ++c;                    else if(nums[a] + nums[b] + nums[c] + nums[d] > target)                        --d;                    else                    {                        result.push_back({nums[a], nums[b], nums[c], nums[d]});                        ++c;                        --d;                    }                }            }        }        sort(result.begin(), result.end());        result.erase(unique(result.begin(), result.end()), result.end());        return result;          }};

2）

class Solution {public:    vector<vector<int>> fourSum(vector<int>& nums, int target) {        vector<vector<int>> result;        if(nums.size() < 4)            return result;        unordered_multimap<int, pair<int, int>> cache;        for(int i = 0; i < nums.size() - 1; ++i)            for(int j = i + 1; j < nums.size(); ++j)                cache.insert(make_pair((nums[i] + nums[j]), make_pair(i, j)));        for(auto i = cache.begin(); i != cache.end(); ++i)        {            int gap = target - i->first;            auto tag = cache.equal_range(gap);            for(auto j = tag.first; j != tag.second; ++j)            {                int a = i->second.first;                int b = i->second.second;                int c = j->second.first;                int d = j->second.second;                if(a != c && a != d && b != c && b != d)                {                    vector<int> vec = {nums[a], nums[b], nums[c], nums[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;    }};