leetcode-Maximum Product Subarray-ZZ

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/