Maximum Subsequence Multiplication

最大连续乘积算法

转载于 2009-06-10 16:02:00 发布 · 49 阅读

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原文链接：http://www.cnblogs.com/graphics/archive/2009/06/10/1500570.html

本文介绍了一种求解整数数组中最大连续乘积的算法。通过维护当前最大正乘积和最小负乘积，该算法能高效地找出数组中最大的连续乘积。

Given an array a contains integers, return
the maximum consecutive multiplication
Example:
int[] a = {2, -2, -3, 4, -5, 6} ;
return 360 (-3 * 4 * -5 * 6)

code

Code
1 // Use two variable max and min to record the max positive multiplication
2 // and min negetive multiplication at current position
3 public long Multiplication(long[] a)
4 {
5     long max = 1; // max positive multiplication at current position
6     long min = 1; // min negetive multiplication at current position
7     long r = 1;   // result, max multiplication totally
8
9     for (int i = 0; i < a.Length; i++)
10     {
11         if (a[i] > 0)
12         {
13             max *= a[i];
14             min = Math.Min(min * a[i], 1);
15         }
16         else if (a[i] == 0)
17         {
18             max = 1;
19             min = 1;
20         }
21         else // i < 0
22         {
23             long temp = max;
24             max = Math.Max(min * a[i], 1);
25             min = temp * a[i];
26         }
27         r = Math.Max(r, max);
28     }
29     return r;
30 }
31

转载于:https://www.cnblogs.com/graphics/archive/2009/06/10/1500570.html