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Find the contiguous subarray within an array (containing at least one number) which has the largest product.
For example, given the array [2,3,-2,4],
the contiguous subarray [2,3] has the largest product = 6.
使用动态规划求解：
新建三个数组suffix_min[n], suffix_max[n], global_max[n]. 分别表示包含最后元素的最小乘积，包含最后元素的最大乘积，和全局最大乘积。
假设已知suffix_min[i-1], suffix_max[i-1],
那么suffix_max[i]是suffix_max[i-1] * A[i], suffix_min[i-1]*A[i], A[i]三者中的最大值，suffix_min[i]是三者中的最小值。
而global_max[i]则是global_max[i-1]和suffix_max[i]中的最大值。
1 int maxProduct(int A[], int n) 2 { 3 if (n <= 0) 4 return n; 5 int *suffix_max = new int[n]; 6 int *suffix_min = new int[n]; 7 int *global_max = new int[n]; 8 int temp_min, temp_max, max_product; 9 10 suffix_max[0] = A[0]; 11 suffix_min[0] = A[0]; 12 global_max[0] = A[0]; 13 for (int i = 1; i < n; i++) 14 { 15 temp_min = A[i] * suffix_min[i - 1]; 16 temp_max = A[i] * suffix_max[i - 1]; 17 suffix_min[i] = min(A[i], min(temp_min, temp_max)); 18 suffix_max[i] = max(A[i], max(temp_min, temp_max)); 19 global_max[i] = max(suffix_max[i], global_max[i - 1]); 20 } 21 max_product = global_max[n - 1]; 22 delete[] suffix_max; 23 delete[] suffix_min; 24 delete[] global_max; 25 26 return max_product; 27 }
对于本题，只需要求出global_max[n]即可，实际上不需要记录全部global_max[0, ..., n -1]
1 int maxProduct(int A[], int n) 2 { 3 if (n <= 0) 4 return n; 5 int temp_min, temp_max; 6 7 int suffix_max = A[0]; 8 int suffix_min = A[0]; 9 int global_max = A[0]; 10 for (int i = 1; i < n; i++) 11 { 12 temp_min = A[i] * suffix_min; 13 temp_max = A[i] * suffix_max; 14 suffix_min = min(A[i], min(temp_min, temp_max)); 15 suffix_max = max(A[i], max(temp_min, temp_max)); 16 global_max = max(suffix_max, global_max); 17 } 18 19 return global_max; 20 }
leetcode 152. Maximum Product Subarray
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原文地址：http://www.cnblogs.com/ym65536/p/4216306.html
