程序员面试题目总结--数组(27)【Two Sum 、3Sum、3Sum Closest 、4Sum】

27、Two Sum

题目：给定一个整数数组，找出和为指定数值的两个数，返回这两个数的下标，要求下标以1为基准

如 Input: numbers={2, 7, 11, 15}, target=9

Output: index1=1, index2=2

原题：Given an array of integers, find two numbers such that they add up to a specific target number.The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please
note that your returned answers (both index1 and index2) are not

zero-based. You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9

Output: index1=1, index2=2

分析：hash法，用一个哈希表，存储每个数对应的下标，复杂度O(n).

#include<iostream>
#include <hash_map>
#include <vector>
#include <algorithm>
using namespace std;
using stdext::hash_map;
vector<int> twoSum(vector<int> &num, int target) 
{
	hash_map<int, int> mapping;
	vector<int> result;
	for (int i = 0; i < num.size(); i++) 
	{
		mapping[num[i]] = i;
	}
	for (int i = 0; i < num.size(); i++) 
	{
		const int gap = target - num[i];
		if (mapping.find(gap) != mapping.end()) 
		{
			result.push_back(i + 1);
			result.push_back(mapping[gap] + 1);
			break;
		}
	}
	return result;
}

扩展1：3Sum

题目：给定一个整型数组，找出这个数组中和为0的三个数，要求找出所有符合条件的数

如 given array S = {-1 0 1 2 -1 -4}.

A solution set is:

(-1, 0, 1)

(-1, -1, 2)

原题：Given an array S of n integers, are there elements a; b; c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

• Elements in a triplet (a; b; c) must be in non-descending order. (ie, a<= b<=c)

• The solution set must not contain duplicate triplets.

For example, given array S = {-1,0,1,2,-1,-4}.

A solution set is:

(-1, 0, 1)

(-1, -1, 2)

分析：先排序，然后左右夹逼，复杂度O(n2)。这个方法可以推广到k-Sum，先排序，然后做k-2次循环，在最内层循环左右夹逼

vector<vector<int> > threeSum(vector<int>& num) 
{
	vector<vector<int> > result;
	if (num.size() < 3) return result;
	sort(num.begin(), num.end());
	const int target = 0;
	vector<int>::iterator last = num.end();
	for (vector<int>::iterator a = num.begin(); a < last; ++a) 
	{
		vector<int>::iterator b = a+1;
		vector<int>::iterator c = last-1;
		while (b < c) 
		{
			if (*a + *b + *c < target)
			{
				++b;
			}
			else if (*a + *b + *c > target)
			{
				--c;
			} 
			else 
			{
				vector<int> tmp;  
				tmp.push_back(*a);  
				tmp.push_back(*b);  
				tmp.push_back(*c);  
				result.push_back(tmp);
				++b;
				--c;
			}
		}
	}
	sort(result.begin(), result.end());
	result.erase(unique(result.begin(), result.end()), result.end());
	return result;
}

扩展2：3Sum Closest

题目：给定一个整型数组，找出和最接近指定数值的三个数，返回三个数的和

原题：Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

For example, given array S = {-1 2 1 -4}, and target = 1.

The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

分析：先排序，然后左右夹逼，复杂度O(n2)

int threeSumClosest(vector<int>& num, int target) {
	int result = 0;
	int min_gap = INT_MAX;
	sort(num.begin(), num.end());
	vector<int>::iterator last = num.end();
	for (vector<int>::iterator a = num.begin(); a != last; ++a) {
		vector<int>::iterator b = a+1;
		vector<int>::iterator c = last-1;
		while (b < c) {
			const int sum = *a + *b + *c;
			const int gap = abs(sum - target);
			if (gap < min_gap) {
				result = sum;
				min_gap = gap;
			}
			if (sum < target) ++b;
			else --c;
		}
	}
	return result;
}

扩展3：4Sum

题目：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)

分析：

方法一：先排序，然后左右夹逼，时间复杂度O(n^3)，空间复杂度O(1)

vector<vector<int> > fourSum(vector<int>& num, int target)
{
	vector<vector<int> > result;
	if (num.size() < 4) return result;
	sort(num.begin(), num.end());
	vector<int>::iterator last = num.end();
	for (vector<int>::iterator a = num.begin(); a < last; ++a) 
	{
		for (vector<int>::iterator b = a+1; b < last; ++b)
		{
			vector<int>::iterator c = b+1;
			vector<int>::iterator d = last-1;
			while (c < d) 
			{
				if (*a + *b + *c + *d < target) 
				{
					++c;
				} else if (*a + *b + *c + *d > target)
				{
					--d;
				} else 
				{
					vector<int> tmp;  
					tmp.push_back(*a);  
					tmp.push_back(*b);  
					tmp.push_back(*c);  
					tmp.push_back(*d);
					result.push_back(tmp);
					++c;
					--d;
				}
			}
		}
	}
	sort(result.begin(), result.end());
	result.erase(unique(result.begin(), result.end()), result.end());
	return result;
}

方法二：map 做缓存。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)

vector<vector<int> > fourSum1(vector<int> &num, int target) {
	vector<vector<int> > result;
	if (num.size() < 4) return result;
	sort(num.begin(), num.end());
	hash_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 vector<pair<int, int> >& vec = cache[key];
			for (size_t k = 0; k < vec.size(); ++k) {
				if (c <= vec[k].second)
					continue; // 有重叠
				vector<int> tmp;  
				tmp.push_back(num[vec[k].first]);  
				tmp.push_back(num[vec[k].second]);  
				tmp.push_back(num[c]);  
				tmp.push_back(num[d]); 
				result.push_back(tmp);
			}
		}
	}
	sort(result.begin(), result.end());
	result.erase(unique(result.begin(), result.end()), result.end());
	return result;
}