leetcode152

class Solution {
public:
    int maxProduct(vector<int>& nums) {
        if(nums.empty())
            return 0;
        if(nums.size() == 1)
            return nums[0];
            
        int maxAll = nums[0];       //global maximum
        int maxLast = nums[0];      //maximum including last element
        int maxCur;                 //maximum including current element
        int minLast = nums[0];      //minimum including current element
        int minCur;                 //minimum including last element
        for(int i = 1; i < nums.size(); i ++)
        {
            maxCur = max(nums[i], max(maxLast*nums[i], minLast*nums[i]));
            minCur = min(nums[i], min(maxLast*nums[i], minLast*nums[i]));
            maxLast = maxCur;
            minLast = minCur;
            maxAll = max(maxAll, maxCur);
        }
        return maxAll;
    }
};

因为nums可能包含负数，因此之前最小的乘积*当前值，有可能成为最大值；而之前最大的乘积*当前值，有可能成为最小值。

因此，每次计算的时候，把目前的最大乘积和最小乘积都保存下来，用于下一次计算。

 

补充一个python的实现：

class Solution:
    def maxProduct(self, nums: 'List[int]') -> 'int':
        n = len(nums)
        if n == 0:
            return 0
        elif n == 1:
            return nums[0]
        maxAll,preMax,preMin = nums[0],nums[0],nums[0]
        curMax,curMin = nums[0],nums[0]
        for i in range(1,n):
            curMax = max(nums[i],max(preMax*nums[i],preMin*nums[i]))
            curMin = min(nums[i],min(preMax*nums[i],preMin*nums[i]))
            preMax,preMin = curMax,curMin
            maxAll = max(maxAll,curMax)
        return maxAll