Optimal Division

最新推荐文章于 2019-05-09 16:09:06 发布

转载最新推荐文章于 2019-05-09 16:09:06 发布 · 112 阅读

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原文链接：http://www.cnblogs.com/amazingzoe/p/7008686.html

本文介绍了一种针对正整数列表的算法，通过不同方式添加括号改变运算优先级来获得最大浮点除法结果，并给出了具体实现代码。

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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.

 1 public class Solution {
 2     public String optimalDivision(int[] nums) {
 3         int n = nums.length;
 4         if (n == 1) return nums[0] + "";
 5         if (n == 2) return nums[0] + "/" + nums[1];
 6         
 7         StringBuilder builder = new StringBuilder();
 8         builder.append(nums[0] + "/(");
 9         for (int i = 1; i < n - 1; i++) {
10             builder.append(nums[i] + "/");
11         }
12         builder.append(nums[n - 1] + ")");
13         
14         return builder.toString();
15     }
16 }

转载于:https://www.cnblogs.com/amazingzoe/p/7008686.html