http://blog.csdn.net/v_july_v/article/details/8701148
假设数组为a[]，直接利用动归来求解，考虑到可能存在负数的情况，我们用Max来表示以a结尾的最大连续子串的乘积值，用Min表示以a结尾的最小的子串的乘积值，那么状态转移方程为：
Max=max{a[i], Max[i-1]*a[i], Min[i-1]*a[i]};
Min=min{a[i], Max[i-1]*a[i], Min[i-1]*a[i]};
初始状态为Max[0]=Min[0]=a[0]。
1 #include <iostream> 2 #include <cmath> 3 #include <algorithm> 4 using namespace std; 5 class Solution { 6 public: 7 int maxProduct(int A[], int n) { 8 int *maxArray = new int[n]; 9 int *minArray = new int[n]; 10 maxArray[0] = minArray[0] = A[0]; 11 int result=maxArray[0]; 12 for (int i = 1; i < n; i++) 13 { 14 maxArray[i] = max(max(maxArray[i-1]*A[i],minArray[i-1]*A[i]),A[i]); 15 minArray[i] = min(min(maxArray[i-1]*A[i],minArray[i-1]*A[i]),A[i]); 16 result = max(result,maxArray[i]); 17 } 18 return result; 19 } 20 }; 21 int main() 22 { 23 Solution s; 24 int n = 4; 25 int a[] = {2,3,-2,4}; 26 cout << s.maxProduct(a,4)<<endl; 27 return 0; 28 }
==============================================================================================
LinkedIn - Maximum Sum/Product Subarray
Maximum Sum Subarray是leetcode原题，跟Gas Station的想法几乎一模一样。解答中用到的结论需要用数学简单地证明一下。
|
1
2
3
4
5
6
7
8
9
10
11
12
|
public int maxSubArray(int[] A) { int sum = 0; int max = Integer.MIN_VALUE; for (int i = 0; i < A.length; i++) { sum += A[i]; if (sum > max) max = sum; if (sum < 0) sum = 0; } return max;} |
Maximum Product Subarray其实只需要不断地记录两个值，max和min。max是到当前为止最大的正product，min是到当前为止最小的负product，或者1。
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
|
public int maxProduct(int[] A) { int x = 1; int max = 1; int min = 1; for (int i = 0; i < A.length; i++) { if (A[i] == 0) { max = 1; min = 1; } else if (A[i] > 0) { max = max * A[i]; min = Math.min(min * A[i], 1); } else { int temp = max; max = Math.max(min * A[i], 1); min = temp * A[i]; } if (max > x) x = max; } return x;} |
http://shepherdyuan.wordpress.com/2014/07/23/linkedin-maximum-sumproduct-subarray/