152. Maximum Product Subarray(中等, 神奇的 swap)

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.

Idea:
https://leetcode.com/problems/maximum-product-subarray/discuss/
keep max(min) value in imax(imin) varable when pointer = i.

swap起到重要作用. 虽然求子数组最大乘积，但最小乘积也需维护，因为： A Truth is: big -> small when (A[i] < 0) small -> big when (A[i] < 0) whatever imax, imin is stored negt or posi in.

人家想法，自个代码：
(O(n)) time, (O(1)) extra space.

// https://leetcode.com/problems/maximum-product-subarray/discuss/
// A = [2,3,-2,4]   return [2,3] = 6
int maxProduct(vector<int>& A) {
	int r = A[0], imax = r, imin = r;
	for (int i = 1; i < A.size(); i++) {
		// A Truth is:
		// big -> small when (A[i] < 0)
		// small -> big when (A[i] < 0)
		// whatever imax, imin is stored negt or posi in.
		if (A[i] < 0)
			swap(imax, imin);

		// keep max(min) value in imax(imin) when pointer = i
		imax = max(A[i], imax * A[i]);

		// 之所以记录 imin,
		// 是因为 if(A[i] < 0) swap(imax, imin);
		// 用到 imin
		imin = min(A[i], imin * A[i]);
		r = max(r, imax);
	}
	return r;
}