[LeetCode] Optimal Division

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.

划分给定的数组，使其相除后的结果最大。要求使括号影响的优先级最简。

数组的首元素一定是被除数。如果要想结果最大，则要求除数最小。

这其实也是一个递归的思想，要求每一个被除数都是最小的。

如果要达成这个目的。只有把每一部分除数当成一个整体来计算。也就是说数组首元素为被除数。其余数顺序相除的结果作为除数。这样的结果最大。

代码如下：

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