Solution

Find the contiguous subarray within an array (containing at least one number) which has the largest product.

Example

For example, given the array[2,3,-2,4], the contiguous subarray[2,3]has the largest product =6.

Solution

因为数组里面有正有负，所以需要同时几率dpMax和dpMin，这样遇到负数的时候，和前面的min相乘将会得到当前的max

public class Solution {
    /*
     * @param nums: An array of integers
     * @return: An integer
     */
    public int maxProduct(int[] nums) {
        // write your code here
        if (nums == null || nums.length == 0) {
            return 0;
        }

        int[] min = new int[nums.length + 1];
        min[0] = 1;
        int[] max = new int[nums.length + 1];
        max[0] = 1;
        int ret = Integer.MIN_VALUE;

        for (int i = 0; i < nums.length; i++) {
            if (nums[i] > 0) {
                max[i + 1] = Math.max(nums[i], nums[i] * max[i]);
                min[i + 1] = Math.min(nums[i], nums[i] * min[i]);
            } else {
                max[i + 1] = Math.max(nums[i], nums[i] * min[i]);
                min[i + 1] = Math.min(nums[i], nums[i] * max[i]);
            }

            ret = Math.max(ret, max[i + 1]);
        }

        return ret;
    }
}