LeetCode OJ：Maximum Product Subarray（子数组最大乘积）

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.

很显然是一个动态规划的问题，找到递归表达式就可以了，考虑到会有负负相乘的问题，所以应该维持一个最大的值以及一个最小的值，递归表达式如下所示：

 maxLocal[i + ] = max(max(maxLocal[i] * nums[i + ], minLocal[i] * nums[i + ]), nums[i + ]);
 minLocal[i + ] = min(min(maxLocal[i] * nums[i + ], minLocal[i] * nums[i + ]), nums[i + ]);
 maxGlobal[i + ] = max(maxGlobal[i], maxLocal[i + ]);

代码如下所示：

 class Solution {
 public:
     int maxProduct(vector<int>& nums) {
         int sz = nums.size();
         if (sz == )return ;
         if (sz == )return nums[];
         int maxP, minP, a, b, ret;
         ret = maxP = minP = a = b = nums[];
         for (int i = ; i < sz; ++i){
             a = max(maxP * nums[i], minP * nums[i]);
             b = min(maxP * nums[i], minP * nums[i]);
             maxP = max(a, nums[i]);
             minP = min(b, nums[i]);
             ret = max(maxP, ret);
         }
         return ret;
     }

 };

java版本的如下所示，注意使用两个中间变量的原因是防止maxVal先直接更新了之后又被minVal所误用：

 public class Solution {
     public int maxProduct(int[] nums) {
         int maxVal, minVal, ret, tmpMax, tmpMin;
         maxVal = minVal = ret = tmpMax = tmpMin = nums[0];
         for(int i = 1; i < nums.length; ++i){
             tmpMax = Math.max(maxVal * nums[i], minVal * nums[i]);
             tmpMin = Math.min(maxVal * nums[i], minVal * nums[i]);
             maxVal = Math.max(tmpMax, nums[i]);
             minVal = Math.min(tmpMin, nums[i]);
             ret = Math.max(maxVal, ret);
         }
         return ret;
     }
 }