【LeetCode】152. Maximum Product Subarray

Maximum Product Subarray

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.

 

解題思路：乘法與加法最大差別在於，當前元素的符號具有全局性的作用。

如果當前元素為負，那么連乘到上個元素的最大乘積，再乘以當前元素，就變成負數，甚至可能成為最小乘積。

同樣，連乘到上個元素的最小乘積如為負，再乘以當前元素，就變成正數，甚至可能成為最大乘積。

 

因此使用動態規划的方法：

記maxLast/minLast為連乘到上個元素的最大/小乘積

記maxCur/minCur為連乘到當前元素的最大/小乘積

記maxAll為全局最大乘積

 

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;
    }
};