Maximum Product Subarray

最新推荐文章于 2021-07-19 16:22:36 发布

转载最新推荐文章于 2021-07-19 16:22:36 发布 · 51 阅读

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原文链接：http://www.cnblogs.com/zengzy/p/5002477.html

本文介绍了一种求解给定数组中具有最大乘积的连续子数组的方法。通过使用动态规划思想，算法能有效地找到最大乘积的子数组。以数组 [2,3,-2,4] 为例，子数组 [2,3] 的乘积最大，等于 6。

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.

class Solution {
public:
    int maxProduct(vector<int>& nums) {
        int numsSize = nums.size();
        int res = 0;
        int preMin = 0;
        int preMax = 0;
        for(int i=0;i<numsSize;i++){
            if(i == 0){
                preMin = nums[0];
                preMax = nums[0];
                res = nums[0];
            }else{
                int tmp = preMax;
                preMax = max(preMin*nums[i],max(preMax*nums[i],nums[i]));
                preMin = min(preMin*nums[i],min(tmp*nums[i],nums[i]));
                res = max(preMax,res);
            }
        }
        return res;
    }
};