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;
}
};