LeetCode "Maximum Product Subarray"

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Intuitively there must a O(n) solution.

First I tried a bottom-up DP solution but it had a TLE:

class Solution {
public:
    int maxProduct(int A[], int n) {
        vector<int> dp;dp.resize(n);
        dp.assign(A, A + n);

        int max = *std::max_element(A, A + n);
        for (size_t len = 2; len <= n; len ++)
        for (int i = 0; i <= n - len; i ++)
        {
            dp[i] *= A[i + len - 1];
            max = std::max(max, dp[i]);
        }

        return max;
    }
};

 So what is O(n) solution then? The idea is to keep track of 2 lines: one for positive and one for negative. Inspired by:

https://oj.leetcode.com/discuss/11923/sharing-my-solution-o-1-space-o-n-running-time

LeetCode "Maximum Product Subarray"

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原文地址：http://www.cnblogs.com/tonix/p/3995924.html