LeetCode 553. Optimal Division （除法划分，算法）

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.

给一组整数，输出如何添加括号和除号使得计算结果最大。

思路：

参考 http://bookshadow.com/weblog/2017/04/16/leetcode-optimal-division/

在不添加任何括号的情况下：

a / b / c / d / ... = a / (b * c * d * ...)

在算式中添加括号会使得被除数和除数的构成发生变化
但无论括号的位置如何，a一定是被除数的一部分，b一定是除数的一部分
原式添加括号方案的最大值，等价于求除数的最小值
因此最优添加括号方案为：

a / (b / c / d / ...) = a * c * d * ... / b

即把更多的数放在被除数位置，除数只剩一个数，此时除数最小，计算结果最大。

因此，不论输入如何，划分的结果都是将第二个元素到最后一个元素用括号括起来。注意输入个数是1时，不添加括号和除号；输入个数是2时，不添加括号。

string optimalDivision(vector<int>& nums) {
int denominatorCnt = nums.size()-1;
string numerator = to_string(nums[0]); //被除数
if(denominatorCnt==0)return numerator; //没有除数的情况
string denominator = ""; //除数
int i;
for(i=1;i<nums.size()-1;i++)
{
denominator+=to_string(nums[i]);
denominator+="/";
}
denominator+=to_string(nums[i]);
if(denominatorCnt>1)denominator = "("+denominator+")";
return numerator+"/"+denominator;
}