算法设计与应用基础：第十二周（2）

152. Maximum Product Subarray

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• Total Accepted: 96023
• Total Submissions: 381825
• Difficulty: Medium
• Contributor: LeetCode

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.

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解题思路:通过多次做题不难得出该问题的状态转移方程A[i]=A[i-1]*nums[i]（A[i]表示包含nums[i]的最大子乘积序列，但是这道题特殊的地方在于它是乘积，也就是说不一定满足以上状态转移方程，分类讨论应该为A[i]=A[i-1]*nums[i]（A[i]为正数时）；A[i]=B[i-1]*nums[i]（A[i]为负数时，B[i]表示包含nums[i]的最小子乘积序列。这样解题就明了了，代码如下： 

 int maxProduct(vector<int>& nums) {    if(nums.empty())        return 0;    else    {        int now_max=nums[0],now_min=nums[0],biggest=nums[0];        for(int i=1;i<nums.size();i++)        {            int temp1=now_min*nums[i],temp2=now_max*nums[i];            if(nums[i]<0)            {                now_max=max(now_min*nums[i],nums[i]);                now_min=min(temp2,nums[i]);                       }            else            {                now_max=max(now_max*nums[i],nums[i]);                now_min=min(now_min*nums[i],nums[i])；                            }            biggest=max(biggest,now_max);        }        return biggest;    }}

 说明：针对方程特性进行了空间复杂度的优化，使得空间复杂度为O(1);