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]。
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
class Solution {
public:
int maxProduct(int A[], int n) {
int *maxArray = new int[n];
int *minArray = new int[n];
maxArray[] = minArray[] = A[];
int result=maxArray[];
for (int i = ; i < n; i++)
{
maxArray[i] = max(max(maxArray[i-]*A[i],minArray[i-]*A[i]),A[i]);
minArray[i] = min(min(maxArray[i-]*A[i],minArray[i-]*A[i]),A[i]);
result = max(result,maxArray[i]);
}
return result;
}
};
int main()
{
Solution s;
int n = ;
int a[] = {,,-,};
cout << s.maxProduct(a,)<<endl;
return ;
}
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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。
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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。
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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/