Given a list of positive integers, the adjacent integers will perform the float division. For example, [2,3,4] -> 2 / 3 / 4.

However, you can add any number of parenthesis at any position to change the priority of operations. You should find out how to add parenthesis to get the maximum result, and return the corresponding expression in string format. Your expression should NOT contain redundant parenthesis.

Example:

Input: [1000,100,10,2]
Output: "1000/(100/10/2)"
Explanation:
1000/(100/10/2) = 1000/((100/10)/2) = 200
However, the bold parenthesis in "1000/((100/10)/2)" are redundant, 
since they don't influence the operation priority. So you should return "1000/(100/10/2)". 

Other cases:
1000/(100/10)/2 = 50
1000/(100/(10/2)) = 50
1000/100/10/2 = 0.5
1000/100/(10/2) = 2

Note:

The length of the input array is [1, 10].
Elements in the given array will be in range [2, 1000].
There is only one optimal division for each test case.

思路：
数学推导，以 b/(c/d) 和 (b/c)/d 为例，
b/(c/d) (b/c)/d = b/c/d
(b*d)/c b/(d*c)
d/c 1/(d*c)
可以看出d/c明显比(1/d)/c大，因为d>1。所以要保证结果最大，需要保证出书了最小，所以除数采用b/c/d的形式会更小。
简单来说，只要后面一加括号，d就会变成乘数。

class Solution {
    public String optimalDivision(int[] nums) {
        StringBuilder builder = new StringBuilder();  
        builder.append(nums[0]);  
        for (int i = 1; i < nums.length; i ++) {  
            if (i == 1 && nums.length > 2) {  
                builder.append("/(").append(nums[i]);  
            } else {  
                builder.append("/").append(nums[i]);  
            }  
        }  
        return nums.length > 2? builder.append(')').toString(): builder.toString();  
    }
}