Given an integer array nums, find the contiguous subarray within an array (containing at least one number) which has the largest product.
Example 1:
Input: [2,3,-2,4]
Output: 6Explanation: [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.
public int maxProduct(int[] A) {
// Loop through the array, each time remember the max and min value for the previous product, the most important thing is to update the max and min value: we have to compare among max * A[i], min * A[i] as well as A[i], since this is product, a negative * negative could be positive.
//最先想到的方比较简单粗暴，就是找出所有的子数组，然后算出每一个子数组的乘积，然后比较找出最大的一个，需要两个for循环，第一个for遍历整个数组，第二个for遍历含有当前数字的子数组，就是按以下顺序找出子数组: [2], [2, 3], [2, 3, -2], [2, 3, -2, 4], [3], [3, -2], [3, -2, 4], [-2], [-2, 4], [4], 我在本地测试的一些数组全部通过，于是兴高采烈的拿到OJ上测试，结果丧心病狂的OJ用一个有15000个数字的数组来测试，然后说我程序的运行时间超过了要求值，我一看我的代码，果然如此，时间复杂度O(n2), 得想办法只用一次循环搞定。我想来想去想不出好方法，于是到网上搜各位大神的解决方法。
//其实这道题最直接的方法就是用DP来做，而且要用两个dp数组，其中f[i]表示子数组[0, i]范围内的最大子数组乘积，g[i]表示子数组[0, i]范围内的最小子数组乘积，初始化时f[0]和g[0]都初始化为nums[0]，其余都初始化为0。那么从数组的第二个数字开始遍历，那么此时的最大值和最小值只会在这三个数字之间产生，即f[i-1]*nums[i]，g[i-1]*nums[i]，和nums[i]。所以我们用三者中的最大值来更新f[i]，用最小值来更新g[i]，然后用f[i]来更新结果res即可
if (A == null || A.length == 0) {
return 0;
}
int max = A[0], min = A[0], result = A[0];
for (int i = 1; i < A.length; i++) {
int temp = max;
max = Math.max(Math.max(max * A[i], min * A[i]), A[i]);
min = Math.min(Math.min(temp * A[i], min * A[i]), A[i]);
if (max > result) {
result = max;
}
}
return result;
}