[刷题防痴呆] 0553 - 最优除法 (Optimal Div

题目地址

https://leetcode.com/problems/optimal-division/

题目描述

553. Optimal Division

You are given an integer array nums. The adjacent integers in nums will perform the float division.

For example, for nums = [2,3,4], we will evaluate the expression "2/3/4".
However, you can add any number of parenthesis at any position to change the priority of operations. You want to add these parentheses such the value of the expression after the evaluation is maximum.

Return the corresponding expression that has the maximum value in string format.

Note: your expression should not contain redundant parenthesis.

 

Example 1:

Input: nums = [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
Example 2:

Input: nums = [2,3,4]
Output: "2/(3/4)"
Example 3:

Input: nums = [2]
Output: "2"

思路

• stringbuffer.
• 数学方法把除数最小.
• a/b/c/d/e的最优解为a/(b/c/d/e)

关键点

• 注意, 边界条件的处理. len为1和2的情况.

代码

• 语言支持：Java

class Solution {
    public String optimalDivision(int[] nums) {
        int n = nums.length;

        if (n == 1) {
            return nums[0] + "";
        }

        if (n == 2) {
            return nums[0] + "/" + nums[1];
        }
        StringBuffer sb = new StringBuffer();
        sb.append(nums[0]);
        sb.append("/(");
        sb.append(nums[1]);
        for (int i = 2; i < n; i++) {
            sb.append("/");
            sb.append(nums[i]);
        }
        sb.append(")");

        return sb.toString();
    }
}