[leetcode] 152. Maximum Product Subarray

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本文探讨了在整数数组中寻找具有最大乘积的连续子数组问题，提供了三种解法：动态规划、优化后的动态规划及双向扫描算法。通过实例解析算法思路，帮助读者理解并掌握最大子数组乘积问题的解决方法。

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1

构建两个动态数组，f[i]记录子数组[0, i]范围内并且一定包含nums[i]数字的最大子数组乘积，g[i]表示子数组[0, i]范围内并且一定包含nums[i]数字的最小子数组乘积。
那么遍历到nums[i]时，最大值和最小值一定在f[i-1]*nums[i]，g[i-1]*nums[i]，和nums[i]中产生。

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

解法2

不用构建新数组来记录，直接使用两个变量mx和mn分别记录当前最大值和当前最小值。

class Solution {
public:
    int maxProduct(vector<int>& nums) {
        int n = nums.size();
        int mx = nums[0],mn = nums[0];
        int res = nums[0];
        for(int i=1;i<n;i++){
            int tmax = mx, tmin = mn;
            mx = max(max(tmax*nums[i], tmin*nums[i]), nums[i]);
            mn = min(min(tmax*nums[i], tmin*nums[i]), nums[i]);
            res = max(mx, res);
        }
        return res;
    }
};

解法3

先从前往后扫一遍，再从后往前扫一遍，遇到0就把p=1.

    int maxProduct(vector<int>& nums) {
        int res = nums[0], prod = 1, n = nums.size();
        for (int i = 0; i < n; ++i) {
            res = max(res, prod *= nums[i]);
            if (nums[i] == 0) prod = 1;
        }
        prod = 1;
        for (int i = n - 1; i >= 0; --i) {
            res = max(res, prod *= nums[i]);
            if (nums[i] == 0) prod = 1;
        }
        return res;
    }