152. Maximum Product Subarray

问题描述:

Given an integer array nums, find the contiguous subarray within an array (containing at least one number) which has the largest product.

Example 1:

Input: [2,3,-2,4]
Output: 6
Explanation: [2,3] has the largest product 6.

Example 2:

Input: [-2,0,-1]
Output: 0
Explanation: The result cannot be 2, because [-2,-1] is not a subarray.

解题思路:

这个我们首先会考虑到累乘的方式来找。

但是由于数组中有负数的存在,所以前面最小的值乘上当前的负数可能就是最大值。

所以我们可以用两个数组来分别存储最小值和最大值。

更新最这个数组时,需要将其与nums[i] 进行比较。

 mx[i] = max(max(prod1, prod2), nums[i]);
 mn[i] = min(min(prod1, prod2), nums[i]);

代码:

class Solution {
public:
    int maxProduct(vector<int>& nums) {
        if(nums.size() == 0)
            return 0;
        if(nums.size() == 1)
            return nums[0];
        vector<int> mx(nums.size());
        vector<int> mn(nums.size());
        mx[0] = nums[0];
        mn[0] = nums[0];
        int ret = mx[0];
        for(int i = 1; i < nums.size(); i++){
            int prod1 = mx[i-1]*nums[i];
            int prod2 = mn[i-1]*nums[i];
            mx[i] = max(max(prod1, prod2), nums[i]);
            mn[i] = min(min(prod1, prod2), nums[i]);
            ret = max(mx[i], ret);
        }
        return ret;
    }
};

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转载自www.cnblogs.com/yaoyudadudu/p/9219389.html