4_Sum

1.问题描述

LeetCode：
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)

2.问题分析

使用一个hashmap缓存两个数之和，两边夹逼。

3.代码

//时间复杂度O(n),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(lasr,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;
	}
};

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(size_t c = 0; c < num.size(); ++c)
		{
			for(size_t d = c+1; d < num.size(); ++d)
			{
				const int key = target - num[a] -num[b];
				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;

					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;
	}
};

4.总结

两个方法，一个是先固定两个数，再使用夹逼；二是使用一个hashmap缓存两个数的和，再使用夹逼。