LeetCode Maximum Product Subarray

最新推荐文章于 2020-02-02 17:32:16 发布

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最新推荐文章于 2020-02-02 17:32:16 发布

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原文链接：http://www.cnblogs.com/lailailai/p/4176446.html

本文提供了一种寻找具有最大乘积的连续子数组的方法。通过动态规划思想，算法能够高效地找到给定数组中乘积最大的子数组。讨论了正负数的影响及如何通过维护当前最大最小值来更新全局最大乘积。

class Solution {
public:
    int maxProduct(int A[], int n) {
        if (A == NULL || n < 1) {
            return INT_MIN;
        }
        int res = INT_MIN;
        int tr = 1;
        for (int i=0; i<n; i++) {
            tr = tr * A[i];
            if (tr > res) {
                res = tr;
            }
            if (tr == 0) {
                tr = 1;
            }
        }
        tr = 1;
        for (int i=n-1; i>=0; i--) {
            tr = tr * A[i];
            if (tr > res) {
                res = tr;
            }
            if (tr == 0) {
                tr = 1;
            }
        }
        return res;
    }
};

更叼的方法，不过速度上没有明显提升

class Solution {
public:
    int maxProduct(int A[], int n) {
        // store the result that is the max we have found so far
        int r = A[0];
    
        // imax/imin stores the max/min product of
        // subarray that ends with the current number A[i]
        for (int i = 1, imax = r, imin = r; i < n; i++) {
            // multiplied by a negative makes big number smaller, small number bigger
            // so we redefine the extremums by swapping them
            if (A[i] < 0)
                swap(imax, imin);
    
            // max/min product for the current number is either the current number itself
            // or the max/min by the previous number times the current one
            imax = max(A[i], imax * A[i]);
            imin = min(A[i], imin * A[i]);
    
            // the newly computed max value is a candidate for our global result
            r = max(r, imax);
        }
        return r;
    }
};

第二轮：

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.

还是中间的方法更好一些

class Solution {
public:
    int maxProduct(int A[], int n) {
        int maxv = A[0];
        int minv = A[0];
        int maxt = A[0];
        for (int i=1; i<n; i++) {
            int mxv = maxv;
            int miv = minv;
            maxv = max(max(mxv * A[i], miv * A[i]), A[i]);
            minv = min(min(miv * A[i], mxv * A[i]), A[i]);
            maxt = max(maxv, maxt);
        }
        return maxt;
    }
};