【C++】【LeetCode】152. Maximum Product Subarray

题目

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.

思路

最笨的办法就是两层for循环遍历数组，时间复杂度O(n^2)；
在讨论区看到一种方法非常巧妙：
循环遍历数组，对于nums[i]，前面的子数组必定有一个最大结果（可能不包括最后一个值），同时记录一个imax，一个imin（这两个值包括了最后一个值）用于考虑是否要把nums[i]纳入结果数组中。如果nums[i]<0，则取前面的最小值和nums[i]相乘，与nums[i]比较出最大值，如果nums[i]>0，则取前面的最大值和nums[i]相乘，与nums[i]比较出最小值。这样，始终记录了当前遍历的结点之前的所有结点组成的数组的最大结果，和能够和后面结点连续的最大值和最小值。

代码

来自于：https://discuss.leetcode.com/topic/4417/possibly-simplest-solution-with-o-n-time-complexity

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;
}