子数组的最大乘积

《编程之美》 2.13

问题：给定一个长度为N的整数数组，只允许用乘法，不能用除法，计算（N-1）个数的组合乘积中最大的一组，并写出算法的时间复杂度。

解法一：把所有N-1个数的组合找出来，分别计算乘积，比较大小。O(N^2)。

int MaxSubProducti(int a[],int n)
{
	assert(n > 1);
	int i,j,k,prod,ret;
	for (i = 0; i < n; i++){
		prod = 1;
		for(j = 0; j < i; j++){
			prod *= a[j];
		}
		for(j = i + 1; j < n; j++){
			prod *= a[j];
		}
		if (i == 0) ret = prod;
		if (prod > ret) ret = prod;
	}
	return ret;
}

int MaxSubProductii(int a[],int n)
{
	assert(n > 1);
	int i,j,prod,ret;
	for(i = 0; i < n; i++){
		prod = 1;
		for(j = 0; j < n; j++){
			prod *= (i == j) ? 1 : a[j];
			if (prod == 0)// terminate early
				break;
		}
		if (i == 0) ret = prod;
		if (prod > ret)	ret = prod;
	}
	return ret;	
}

解法二：利用两个数组s和s分别表示从数组a的两端开始计算的乘积，s[i] = a[0]*a[1]*...*a[i-1]，t[j] = a[j]*...*a[n-1]。那么p[i] = s[i-1] * t[i+1]。时间复杂度为O(N)。

int MaxSubProductiii(int a[],int n)
{
	assert(n > 1);
	int *s = (int *)malloc(n*sizeof(int));
	int *t = (int *)malloc(n*sizeof(int));
	int i,j,lowprod,highprod,maxprod,prod;
	lowprod = 1;
	highprod = 1;
	assert(s != NULL && t != NULL);
	
	for(i = 0; i < n; i++){
		j = n - i - 1;
		lowprod *= a[i];
		highprod *= a[j];
		s[i] = lowprod;
		t[j] = highprod;
	}
	maxprod = (s[n-2] > t[1]) ? s[n-2] : t[1];
	for(i = 1; i < n - 1; i ++){
		prod = s[i-1] * t[i+1];
		if (prod > maxprod) maxprod = prod;
	}
	free(s);
	free(t); 
	return maxprod;
}

解法三：统计数组中正数、负数和零的个数，记录最小的正数、负数以及最大的负数。

1，如果零的个数n_zero大于1，那么最大乘积为0；

2，如果零的个数n_zero为1：

(1)，如果负数的个数n_neg为奇数，那么最大乘积为0；

(2)，如果负数的个数n_neg为偶数，那么最大乘积计算式扣除a[zeroi]；

3，如果零的个数n_zero为0：

(1)，如果负数的个数n_neg为奇数，那么最大乘积计算式扣除a[maxnegi]；

(2)，如果负数的个数n_neg为偶数：

1> 如果正数的个数n_pos为零，那么最大乘积计算式扣除a[minnegi]；

2> 如果正数的个数n_pos大于零，那么最大乘积计算式扣除a[minposi]；

int MaxSubProductiv(int *a, int n)
{
	int n_zero = 0; // count of zero 
	int n_neg = 0;  // count of negative integer
	int n_pos = 0;
	int maxneg = 0; // maximum negative integer 
	int minpos = 0; // minimum positive integer 
	int maxnegi = 0; // index of maxneg
	int minposi = 0; // index of minimum positive integer
	int zeroi = 0;   // index of zero
	int minneg = 0;
	int minnegi = 0;
	int out = -1;    // index of the val to be omitted 
	                 // when calculating result product
	int max = 1;
	
	/* collect some statistic info in one loop */
	for(int i = 0; i < n; i++)
	{
		if(a[i] < 0)  //a[i] is negative
		{
			n_neg++;
			if(maxneg == 0)
			{
				maxneg = a[i];
				maxnegi = i;
			}
			else if(maxneg < a[i])
			{
				maxneg = a[i];
				maxnegi = i;
			}
			
			if (minneg == 0) {
				minneg = a[i];
				minnegi = i;
			}
			else if (minneg > a[i]) {
				minneg = a[i];
				minnegi = i;
			}
		}
		else if(a[i] == 0) // a[i] is zero
		{
			zeroi = i;
			if(++n_zero > 1) return 0;
		}
		else // a[i] is positive
		{
			n_pos++;
			if(minpos == 0)
			{
				minpos = a[i];
				minposi = i;
			}
			else if(minpos > a[i])
			{
				minpos = a[i];
				minposi = i;
			}
		}
	}
 
 
	if((n_zero == 1) && (n_neg & 0x1))
	{
		// one zero and odd negatives
		return 0;
	}
	else if((n_zero == 1) && !(n_neg & 0x1))
	{
		// one zero and even negatives
		printf("one zero and even negatives\n");
		out = zeroi;
	}
	else if((n_zero == 0) && (n_neg & 0x1))
	{
		// no zero and odd negatives
		printf("no zero and odd negatives\n");
		out = maxnegi;
	}
	else if((n_zero == 0) && !(n_neg & 0x1))
	{
		// no zero and even negatives
		printf("no zero and even negatives\n");
		if (n_pos == 0){
			out = minnegi;
		}
		else {
			out = minposi;
		}
	}
	
	for(int i = 0; i < n; i++)
	{
		max *= (i == out) ? 1 : a[i];
	}
	return max;
}

int MaxSubProductv(int *a, int n)
{
	int n_zero = 0; // count of zero 
	int n_neg = 0;  // count of negative integer
	int n_pos = 0;  // count of positive integer
	int maxneg = 0; // maximum negative integer 
	int minpos = 0; // minimum positive integer 
	int maxnegi = 0; // index of maxneg
	int minposi = 0; // index of minimum positive integer
	int zeroi = 0;   // index of zero
	
	int minneg = 0;
	int minnegi = 0;
	
	int out = -1;         // index of the val to be omitted 
	                 // when calculating  product
	int max = 1;
	
	for(int i = 0; i < n; i++)
	{
		if(a[i] < 0)  //a[i] is negative
		{
			n_neg++;
			if(maxneg == 0)
			{
				maxneg = a[i];
				maxnegi = i;
			}
			else if(maxneg < a[i])
			{
				maxneg = a[i];
				maxnegi = i;
			}
			
			if (minneg == 0){
				minneg = a[i];
				minnegi = i; 
			}
			else {
				if (minneg > a[i]){
					minneg = a[i];
					minnegi = i;
				}
			} 
		}
		else if(a[i] == 0) // a[i] is zero
		{
			zeroi = i;
			if(++n_zero > 1) return 0;
		}
		else // a[i] is positive
		{
			n_pos++;
			if(minpos == 0)
			{
				minpos = a[i];
				minposi = i;
			}
			else if(minpos > a[i])
			{
				minpos = a[i];
				minposi = i;
			}
		}
	}
 
 	if (n_zero){
 		printf("zeros.\n");
 		if (n_neg & 1)
 			return 0;
 		else
 			out = zeroi;
 	}
 	else {
 		printf("no zero.\n");
 		if (n_neg & 1)
 			out = maxnegi;
 		else if (n_pos == 0){
 			out = minnegi;
 		}
 		else {
 			out = minposi;
 		}
 	}
		
	for(int i = 0; i < n; i++)
	{
		max *= (i == out) ? 1 : a[i];
	}
	return max;
}

测试程序：

#include <stdio.h>
#include <assert.h>
#include <time.h>
#include <stdlib.h>


int MaxSubProducti(int a[],int n)
{
	assert(n > 1);
	int i,j,k,prod,ret;
	for (i = 0; i < n; i++){
		prod = 1;
		for(j = 0; j < i; j++){
			prod *= a[j];
		}
		for(j = i + 1; j < n; j++){
			prod *= a[j];
		}
		if (i == 0) ret = prod;
		if (prod > ret) ret = prod;
	}
	return ret;
}

int MaxSubProductii(int a[],int n)
{
	assert(n > 1);
	int i,j,prod,ret;
	for(i = 0; i < n; i++){
		prod = 1;
		for(j = 0; j < n; j++){
			prod *= (i == j) ? 1 : a[j];
			if (prod == 0)// terminate early
				break;
		}
		if (i == 0) ret = prod;
		if (prod > ret)	ret = prod;
	}
	return ret;	
}

int MaxSubProductiii(int a[],int n)
{
	assert(n > 1);
	int *s = (int *)malloc(n*sizeof(int));
	int *t = (int *)malloc(n*sizeof(int));
	int i,j,lowprod,highprod,maxprod,prod;
	lowprod = 1;
	highprod = 1;
	assert(s != NULL && t != NULL);
	
	for(i = 0; i < n; i++){
		j = n - i - 1;
		lowprod *= a[i];
		highprod *= a[j];
		s[i] = lowprod;
		t[j] = highprod;
	}
	maxprod = (s[n-2] > t[1]) ? s[n-2] : t[1];
	for(i = 1; i < n - 1; i ++){
		prod = s[i-1] * t[i+1];
		if (prod > maxprod) maxprod = prod;
	}
	free(s);
	free(t); 
	return maxprod;
}

int MaxSubProductiv(int *a, int n)
{
	int n_zero = 0; // count of zero 
	int n_neg = 0;  // count of negative integer
	int n_pos = 0;
	int maxneg = 0; // maximum negative integer 
	int minpos = 0; // minimum positive integer 
	int maxnegi = 0; // index of maxneg
	int minposi = 0; // index of minimum positive integer
	int zeroi = 0;   // index of zero
	int minneg = 0;
	int minnegi = 0;
	int out = -1;    // index of the val to be omitted 
	                 // when calculating result product
	int max = 1;
	
	/* collect some statistic info in one loop */
	for(int i = 0; i < n; i++)
	{
		if(a[i] < 0)  //a[i] is negative
		{
			n_neg++;
			if(maxneg == 0)
			{
				maxneg = a[i];
				maxnegi = i;
			}
			else if(maxneg < a[i])
			{
				maxneg = a[i];
				maxnegi = i;
			}
			
			if (minneg == 0) {
				minneg = a[i];
				minnegi = i;
			}
			else if (minneg > a[i]) {
				minneg = a[i];
				minnegi = i;
			}
		}
		else if(a[i] == 0) // a[i] is zero
		{
			zeroi = i;
			if(++n_zero > 1) return 0;
		}
		else // a[i] is positive
		{
			n_pos++;
			if(minpos == 0)
			{
				minpos = a[i];
				minposi = i;
			}
			else if(minpos > a[i])
			{
				minpos = a[i];
				minposi = i;
			}
		}
	}
 
 
	if((n_zero == 1) && (n_neg & 0x1))
	{
		// one zero and odd negatives
		return 0;
	}
	else if((n_zero == 1) && !(n_neg & 0x1))
	{
		// one zero and even negatives
		printf("one zero and even negatives\n");
		out = zeroi;
	}
	else if((n_zero == 0) && (n_neg & 0x1))
	{
		// no zero and odd negatives
		printf("no zero and odd negatives\n");
		out = maxnegi;
	}
	else if((n_zero == 0) && !(n_neg & 0x1))
	{
		// no zero and even negatives
		printf("no zero and even negatives\n");
		if (n_pos == 0){
			out = minnegi;
		}
		else {
			out = minposi;
		}
	}
	
	for(int i = 0; i < n; i++)
	{
		max *= (i == out) ? 1 : a[i];
	}
	return max;
}

int MaxSubProductv(int *a, int n)
{
	int n_zero = 0; // count of zero 
	int n_neg = 0;  // count of negative integer
	int n_pos = 0;  // count of positive integer
	int maxneg = 0; // maximum negative integer 
	int minpos = 0; // minimum positive integer 
	int maxnegi = 0; // index of maxneg
	int minposi = 0; // index of minimum positive integer
	int zeroi = 0;   // index of zero
	
	int minneg = 0;
	int minnegi = 0;
	
	int out = -1;         // index of the val to be omitted 
	                 // when calculating  product
	int max = 1;
	
	for(int i = 0; i < n; i++)
	{
		if(a[i] < 0)  //a[i] is negative
		{
			n_neg++;
			if(maxneg == 0)
			{
				maxneg = a[i];
				maxnegi = i;
			}
			else if(maxneg < a[i])
			{
				maxneg = a[i];
				maxnegi = i;
			}
			
			if (minneg == 0){
				minneg = a[i];
				minnegi = i; 
			}
			else {
				if (minneg > a[i]){
					minneg = a[i];
					minnegi = i;
				}
			} 
		}
		else if(a[i] == 0) // a[i] is zero
		{
			zeroi = i;
			if(++n_zero > 1) return 0;
		}
		else // a[i] is positive
		{
			n_pos++;
			if(minpos == 0)
			{
				minpos = a[i];
				minposi = i;
			}
			else if(minpos > a[i])
			{
				minpos = a[i];
				minposi = i;
			}
		}
	}
 
 	if (n_zero){
 		printf("zeros.\n");
 		if (n_neg & 1)
 			return 0;
 		else
 			out = zeroi;
 	}
 	else {
 		printf("no zero.\n");
 		if (n_neg & 1)
 			out = maxnegi;
 		else if (n_pos == 0){
 			out = minnegi;
 		}
 		else {
 			out = minposi;
 		}
 	}
		
	for(int i = 0; i < n; i++)
	{
		max *= (i == out) ? 1 : a[i];
	}
	return max;
}


void show(int a[],int n)
{
	for(int i = 0; i < n; i++){
		printf("%-3d",a[i]);
	}
	printf("\n");
}
int main()
{
	int a[6];
	int N = 6;
	srand(time(NULL));
	for(int i = 0; i < N; i++){
		a[i] = rand() % 10 ; 
		if (a[i] & 0x1)
			a[i] = -a[i];
	}
	show(a,N);
	printf("Max Sub Product of N-1 elements: %d .\n",MaxSubProducti(a,N));
	printf("Max Sub Product of N-1 elements: %d .\n",MaxSubProductii(a,N));
	printf("Max Sub Product of N-1 elements: %d .\n",MaxSubProductiii(a,N));
	printf("Max Sub Product of N-1 elements: %d .\n",MaxSubProductiv(a,N));
	printf("Max Sub Product of N-1 elements: %d .\n",MaxSubProductv(a,N));
} 

测试输出：

2  -5 -5 -5 -7 2  
Max Sub Product of N-1 elements: 1750 .
Max Sub Product of N-1 elements: 1750 .
Max Sub Product of N-1 elements: 1750 .
no zero and even negatives
Max Sub Product of N-1 elements: 1750 .
no zero.
Max Sub Product of N-1 elements: 1750 .

REF:

1，编程之美 2.13