553. Optimal Division

553. 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 themaximum 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) = 200However, 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 = 501000/(100/(10/2)) = 501000/100/10/2 = 0.51000/100/(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.

        这道题给一个数组，求不改变数列顺序的条件下，这些数相除怎样结果最大，结果用字符串的方式输出。其实比较简单，第一个数是必须要除第二个数的，当除数不变只能被除数减小，所以就是第一个数除以后边所有数相除的结果。其值等于第一个数乘以除了第二个数以外的所有数然后除以第二个数(nums[0]*nums[2]*……*nums[nums.size()-1])/nums[1]。这样就只需考虑给定数组长度变化是字符串方面的差异就可以了。

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