leetcode题解（五十二）：152. Maximum Product Subarray

最新推荐文章于 2024-02-19 10:45:53 发布

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最新推荐文章于 2024-02-19 10:45:53 发布

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博客围绕求最大子数组的积展开，指出此类问题通常用动态规划解决。由于是求乘积，不仅要记录最大值，还需记录最小值，并给出两个示例，分别展示输入数组及对应的最大子数组积的输出结果。

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题意：求最大的子数组的积
一般这种题都是动态规划做的
因为是乘积，所以不仅要记下最大，还要记下最小

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.

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;
}