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:

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

劃分給定的數組，使其相除後的結果最大。要求使括號影響的優先級最簡。

數組的首元素一定是被除數。如果要想結果最大，則要求除數最小。

這其實也是一個遞歸的思想，要求每一個被除數都是最小的。

如果要達成這個目的。只有把每一部分除數當成一個整體來計算。也就是說數組首元素為被除數。其余數順序相除的結果作為除數。這樣的結果最大。

代碼如下：

class Solution {
public
 
:
    string optimalDivision(vector<int>& nums) {
        string str;
        if (nums.empty())
            return str;
        str = to_string(nums[0]);
        if (nums.size() == 1)
            return str;
        if (nums.size() == 2)
            return str + "/" + to_string(nums[1]);
        str += "/(" + to_string(nums[1]);
        for (int i = 2; i < nums.size(); i++) 
            str += "/" + to_string(nums[i]);
        str += ")";
        return str;
    }
};
// 4 ms

[LeetCode] Optimal Division