Leetcode 4Sum 性能瓶颈之容器析构

最新推荐文章于 2025-09-08 18:09:18 发布

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#leetcode #性能 #析构

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本文通过分析LeetCode上的4Sum问题，提出了一种预处理方法并详细探讨了其时间复杂度。通过对不同数据规模的实验对比，揭示了容器析构在实际运行中对性能的影响。

最近做Leetcode上的4Sum，自己想到了一个不太好的方法，刚刚好可以过，代码如下：

#include <iostream>
#include <string>
#include <vector>
#include <stdlib.h>
#include <map>
#include <algorithm>
#include <set>
#include <time.h>
using namespace std;

struct From
{
	bool used;
	int m_a;
	int m_b;
	From(int a, int b) :m_a(a), m_b(b), used(false)
	{
	}
};

bool check(const From& l, const From& r)
{
	if (l.m_a == r.m_a || l.m_a == r.m_b ||
		l.m_b == r.m_a || l.m_b == r.m_b)
	{
		// 说明有重复
		return false;
	}
	return true;
}

class Solution {
private:

public:
	vector<vector<int> > fourSum(vector<int> &num, int target) {
		clock_t start, finish;
		clock_t start0, finish0;
		double duration;

		start0 = clock();

		start = clock();

		//预处理
		multimap<int, From> newNums;

		for (int i = 0; i < num.size(); ++i)
		{
			for (int j = i + 1; j < num.size(); ++j)
			{
				if (num[i] <= num[j])
				{
					newNums.insert(make_pair(num[i] + num[j], From(i, j)));
				}
				else
				{
					newNums.insert(make_pair(num[i] + num[j], From(j, i)));
				}
			}
		}

		finish = clock();
		duration = (double)(finish - start) / CLOCKS_PER_SEC;
		printf("%f seconds\n", duration);

		start = clock();

		vector<vector<int>> ans;
		set <vector<int>> duplicate;

		auto l = newNums.begin();

		for (auto i : newNums)
		{
			auto ret = newNums.equal_range(target - i.first);
			for (auto j = ret.first; j != ret.second; ++j)
			{
				if (j->second.used)
				{
					continue;
				}
				if (check(i.second, j->second))
				{
					vector<int> solution;
					solution.push_back(num[i.second.m_a]);
					solution.push_back(num[i.second.m_b]);
					solution.push_back(num[j->second.m_a]);
					solution.push_back(num[j->second.m_b]);
					sort(solution.begin(), solution.end());
					if (duplicate.find(solution) == duplicate.end())
					{
						duplicate.insert(solution);
						ans.push_back(solution);
					}
				}
			}

			i.second.used = true;
		}

		finish = clock();
		duration = (double)(finish - start) / CLOCKS_PER_SEC;
		printf("%f seconds\n", duration);

		finish0 = clock();
		duration = (double)(finish0 - start0) / CLOCKS_PER_SEC;
		printf("Function Run : %f seconds\n", duration);

		return ans;
	}
};

我自己分析了一下复杂度，预处理是O(n^2*log(n^2))，newNums中的个数应该为O(n^2)，之后在newNums中找找满足条件的结果，这段代码的复杂度应为O(N^2*log(N^2))*log(k)

前面的O(N^2*log(N^2))是最外面的for循环加上在multimap中查找元素的消耗，后面的log(k)，k是去重set中的个数，k最坏情况可能有O(n^2)*O(n^2)个。

所以最后我这方法的复杂度应该是O(N*N*logN*logN) PS：希望没有算错

按理说在大数据量的情况下比O(N^3)的算法要快，我觉得应该是Leetcode的数据量比较小的缘故，所以构造multimap和set浪费了不少时间而体现不出来优势。我就自己写了个测试用例来看看到底这个算法是慢在哪里。测试用例如下：

int main()
{
	clock_t start, finish;
	double duration;
	string str;
	vector<int> num;// = { 91277418, 66271374, 38763793, 4092006, 11415077, 60468277, 1122637, 72398035, -62267800, 22082642, 60359529, -16540633, 92671879, -64462734, -55855043, -40899846, 88007957, -57387813, -49552230, -96789394, 18318594, -3246760, -44346548, -21370279, 42493875, 25185969, 83216261, -70078020, -53687927, -76072023, -65863359, -61708176, -29175835, 85675811, -80575807, -92211746, 44755622, -23368379, 23619674, -749263, -40707953, -68966953, 72694581, -52328726, -78618474, 40958224, -2921736, -55902268, -74278762, 63342010, 29076029, 58781716, 56045007, -67966567, -79405127, -45778231, -47167435, 1586413, -58822903, -51277270, 87348634, -86955956, -47418266, 74884315, -36952674, -29067969, -98812826, -44893101, -22516153, -34522513, 34091871, -79583480, 47562301, 6154068, 87601405, -48859327, -2183204, 17736781, 31189878, -23814871, -35880166, 39204002, 93248899, -42067196, -49473145, -75235452, -61923200, 64824322, -88505198, 20903451, -80926102, 56089387, -58094433, 37743524, -71480010, -14975982, 19473982, 47085913, -90793462, -33520678, 70775566, -76347995, -16091435, 94700640, 17183454, 85735982, 90399615, -86251609, -68167910, -95327478, 90586275, -99524469, 16999817, 27815883, -88279865, 53092631, 75125438, 44270568, -23129316, -846252, -59608044, 90938699, 80923976, 3534451, 6218186, 41256179, -9165388, -11897463, 92423776, -38991231, -6082654, 92275443, 74040861, 77457712, -80549965, -42515693, 69918944, -95198414, 15677446, -52451179, -50111167, -23732840, 39520751, -90474508, -27860023, 65164540, 26582346, -20183515, 99018741, -2826130, -28461563, -24759460, -83828963, -1739800, 71207113, 26434787, 52931083, -33111208, 38314304, -29429107, -5567826, -5149750, 9582750, 85289753, 75490866, -93202942, -85974081, 7365682, -42953023, 21825824, 68329208, -87994788, 3460985, 18744871, -49724457, -12982362, -47800372, 39958829, -95981751, -71017359, -18397211, 27941418, -34699076, 74174334, 96928957, 44328607, 49293516, -39034828, 5945763, -47046163, 10986423, 63478877, 30677010, -21202664, -86235407, 3164123, 8956697, -9003909, -18929014, -73824245 };
	for (int i = 0; i < 1500; ++i)
	{
		num.push_back(rand());
	}
	Solution s;
	start = clock();
	auto ans = s.fourSum(num, rand());
	finish = clock();
	duration = (double)(finish - start) / CLOCKS_PER_SEC;
	printf("Total %d : %f seconds\n", ans.size(), duration);
}

上面被注释掉的是从Leetcode里抓出来的一组解，他的N只有200，确实数据量很小，我就想测测当N等于1500时候的时间分布，用VS2013 Release下运行结果如下：