【刷题-LeetCode】152 Maximum Product Subarray

152. Maximum Product Subarray

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.

解法 动态规划。需要设置两个dp数组，分别保存到位置i的最大和最小连续乘积，这是因为最小的乘积可能是负的，如果再乘上一个负数会变成比较大数

\[\mathrm{dp\_max}[0] = \mathrm{dp\_min}[0] = \mathrm{nums}[0]\\ \mathrm{dp\_max[i]} = \max\{\mathrm{nums}[i], \mathrm{dp\_max}[i-1]*\mathrm{nums}[i], \mathrm{dp\_min}[i-1]*\mathrm{nums}[i]\}\\ \mathrm{dp\_min[i]} = \min\{\mathrm{nums}[i], \mathrm{dp\_max}[i-1]*\mathrm{nums}[i], \mathrm{dp\_min}[i-1]*\mathrm{nums}[i]\} \]

对于数组dp_max和dp_min，每次更新只用两个连续的数，因此可以将空间复杂度优化为\(O(1)\)

class Solution {
public:
    int maxProduct(vector<int>& nums) {
        int pre_f = nums[0], pre_g = nums[0];
        int res = pre_f;
        for(int i = 1; i < nums.size(); ++i){
            int tmp1 = max(nums[i], max(pre_f*nums[i], pre_g*nums[i]));
            int tmp2 = min(nums[i], min(pre_f*nums[i], pre_g*nums[i]));
            res = max(res, tmp1);
            pre_f = tmp1;
            pre_g = tmp2;
        }
        return res;
    }
};