Maximum Product Subarray

最新推荐文章于 2021-07-19 16:22:36 发布

原创最新推荐文章于 2021-07-19 16:22:36 发布 · 296 阅读

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本文探讨了如何寻找数组中乘积最大的连续子数组问题，给出了一种有效的算法实现，并逐步优化了初始解决方案。

Find the contiguous subarray within an array (containing at least one number) which has the largest product.

For example, given the array [2,3,-2,4],
the contiguous subarray [2,3] has the largest product = 6.

class Solution 
{
public:
	int maxProduct(int A[], int n) 
	{
		int startPos = -1, startNeg = -1;
		int resPos = 1, resNeg = 1, res = INT_MIN;
		for (int i = 0; i < n; i++)
		{
			res = max(res, A[i]);
			if (0 == A[i])
			{
				resPos = 1;
				resNeg = 1;
				startPos = -1;
				startNeg = -1;
				continue;
			}
			if (A[i] > 0 && -1 == startPos)
				startPos = i;
			if (A[i] < 0 && -1 == startNeg)
				startNeg = i;
			if (-1 != startPos)
				resPos *= A[i];
			if (-1 != startNeg)
				resNeg *= A[i];
			if (startPos != -1)
				res = max(res, resPos);
			if (startNeg != -1)
				res = max(res, resNeg);
			if (startNeg != -1 && A[startNeg] != resNeg)
				res = max(res, resNeg/A[startNeg]);

		}	
		return res;
	}
};

通过后发现可以稍微改进一下，没必要记录第一个正数位置。第一个负数是影响整个结果的关键，除去它或保留它都有可能产生 Maximum Product

class Solution {
public:
	int maxProduct(int A[], int n) 
	{
		if (0 == n) return 0;
		int res = A[0];
		int resBegin=1, resNeg=1, startNeg=-1;
		for (int i = 0; i < n; i++)
		{
			res = max(res, A[i]);
			if (0 == A[i])
			{
				resBegin = 1;
				resNeg = 1;
				startNeg = -1;
				continue;
			}
			resBegin *= A[i];
			res = max(res,  resBegin);
			if (-1 == startNeg && A[i] < 0)
				startNeg = i;
			if (-1 != startNeg)
			{
				resNeg *= A[i];
				res = max(res, resNeg);
				if (resNeg !=  A[startNeg])
					res = max(res, resNeg/A[startNeg]);
			}							
		}
		return res;
	}					   
};

再次修改简化后的，发现只要i和start negative不是同一个位置，就可以去掉多余判断

class Solution 
{
public:
	int maxProduct(int A[], int n) 
	{
		if (0 == n) return 0;
		int res = A[0];
		int resBegin=1, resNeg=1, startNeg=-1;
		for (int i = 0; i < n; i++)
		{
			res = max(res, A[i]);
			if (0 == A[i])
			{
				resBegin = 1;
				resNeg = 1;
				startNeg = -1;
				continue;
			}
			resBegin *= A[i];
			res = max(res,  resBegin);
			if(-1 != startNeg)
			{
				resNeg *= A[i];
				res = max(res, max(resNeg, resNeg/A[startNeg]));
			}
			if (-1 == startNeg && A[i] < 0)
				startNeg = i;
		}	
		return res;
	}					   
};