leetcode_16题——3Sum Closest（两个指针）

3Sum Closest

Total Accepted: 38536 Total Submissions: 143223My Submissions

Question Solution

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).

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Array Two Pointers

开始的时候想半天，没想到啥好办法，开始想的先计算两个的与target的距离的最小，确定了之后，再计算三个的，发现这样做算法上就是错误的。

int threeSumClosest(vector<int>& nums, int target) {
	int len_nums=nums.size();
	int x=0;
	int y=1;
	int best_two_sum=abs(nums[0]+nums[1]-target);
	for(int i=0;i<len_nums-1;i++)
		for(int j=i+1;j<len_nums;j++)
		{
			if(abs(nums[i]+nums[j]-target)<best_two_sum)
			{
				x=i;
				y=j;
				best_two_sum=abs(nums[i]+nums[j]-target);
			}
		}

	int z=0;
	int flag=0;
	int last_result;
	for(int i=0;i<len_nums;i++)
	{
		if(i!=x&&i!=y&&flag==0)
		{z=i;last_result=abs(nums[i]+nums[x]+nums[y]-target);flag=1;}
		if(i!=x&&i!=y&&flag==1)
		{
			if(abs(nums[i]+nums[x]+nums[y]-target)<last_result)
			{
				z=i;
				last_result=abs(nums[i]+nums[x]+nums[y]-target);
			}
		}
	}
	return nums[x]+nums[y]+nums[z];
}
*/
//O(n^3)时间复杂度太高了
/*
int threeSumClosest(vector<int>& nums, int target)
{
	int len_nums=nums.size();
	int result=abs(nums[0]+nums[1]+nums[2]-target);
	int l_x=0,l_y=1,l_z=2;
	for(int y=1;y<len_nums-1;y++)
	{
		for(int x=0;x<y;x++)
			for(int z=y+1;z<len_nums;z++)
			{
				if(abs(nums[x]+nums[y]+nums[z]-target)<result)
				{
					l_x=x;l_y=y;l_z=z;
					result=abs(nums[x]+nums[y]+nums[z]-target);
				}
			}
	}
	return nums[l_x]+nums[l_y]+nums[l_z];
}

　　接下来想到干脆直接采用穷举法，试下，也就是时间复杂度为O（n^3）发现不能AC

int threeSumClosest(vector<int>& nums, int target)
{
	int len_nums=nums.size();
	int result=abs(nums[0]+nums[1]+nums[2]-target);
	int l_x=0,l_y=1,l_z=2;
	for(int y=1;y<len_nums-1;y++)
	{
		for(int x=0;x<y;x++)
			for(int z=y+1;z<len_nums;z++)
			{
				if(abs(nums[x]+nums[y]+nums[z]-target)<result)
				{
					l_x=x;l_y=y;l_z=z;
					result=abs(nums[x]+nums[y]+nums[z]-target);
				}
			}
	}
	return nums[l_x]+nums[l_y]+nums[l_z];
}

　　   再后来在网上看了下别人的算法，发现有一种还可以，就是在开始的时候就进行排序，这个是O（nlog n），接着枚举一个数，再设立两个指针，一个左指针从0到这个数的位置，一个右指针从最后到这个数的位置，然后当sum<target时将左指针右移，当sum>target时，将右指针向左移，

值得注意的是，在这个移的过程中，当sum<target时，若右移后发现又变成了sum>target，则要将之前的值记录下来，进行比较，另外一个指针也是

同样的道理。这个时间复杂度是O（n^2），

总的为O（nlog n）+O（n^2）=O(n^2)

int threeSumClosest(vector<int>& nums, int target)
{
	sort(nums.begin(),nums.end());
	int len_nums=nums.size();
	int sum;
	int sum_reault;

	int last_sum=abs(nums[0]+nums[1]+nums[2]-target);
	sum_reault=nums[0]+nums[1]+nums[2];

	for(int y=1;y<len_nums-1;y++)
	{
		int x=0;
		int z=len_nums-1;
		while(x<y&&z>y)
		{
			if(nums[x]+nums[y]+nums[z]>target)
			{
				z--;
				if(z>y&&nums[x]+nums[y]+nums[z]<target)
				{
					sum=nums[x]+nums[y]+nums[z+1];
					if(abs(sum-target)<last_sum)
					{
						last_sum=abs(sum-target);
						sum_reault=sum;
					}
				}
			}
			else if(nums[x]+nums[y]+nums[z]<target)
			{
				x++;
				if(x<y&&nums[x]+nums[y]+nums[z]>target)
				{
					sum=nums[x-1]+nums[y]+nums[z];
					if(abs(sum-target)<last_sum)
					{
						last_sum=abs(sum-target);
						sum_reault=sum;
					}
				}
			}
			else
				return target;
		}

		if(x==y)
		{
			x-=1;
			sum=nums[x]+nums[y]+nums[z];
			if(abs(sum-target)<last_sum)
			{
				last_sum=abs(sum-target);
				sum_reault=sum;

			}
		}
		if(z==y)
		{
			z+=1;
			sum=nums[x]+nums[y]+nums[z];
			if(abs(sum-target)<last_sum)
			{
				last_sum=abs(sum-target);
				sum_reault=sum;
			}
		}
	}
	return sum_reault;
}

int main()
{
	vector<int> vec;
	vec.push_back(0);
	vec.push_back(2);
	vec.push_back(1);
	vec.push_back(-3);

	cout<<threeSumClosest(vec,1)<<endl;

}

　　

时间： 2024-10-27 05:34:52

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