leetcode 553. Optimal Division

最新推荐文章于 2019-05-09 16:09:06 发布

qq_27262609

最新推荐文章于 2019-05-09 16:09:06 发布

阅读量876

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/qq_27262609/article/details/70208042

本文介绍了一种算法，用于给定正整数列表时通过合理添加括号改变运算优先级来获得最大结果的表达式。文章提供了一个具体的例子，展示了如何处理不同长度的输入数组，并给出了对应的解决方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

public class Solution {
    public String optimalDivision(int[] nums) {
        String str="";
        if(nums.length==1)return nums[0]+"";
        if(nums.length==2)return nums[0]+"/"+nums[1];
        for(int i=1;i<nums.length-1;i++){
            str=str+nums[i]+"/";
        }
        if(nums.length>1)str=str+nums[nums.length-1];
        return nums[0]+"/"+"("+str+")";
    }
}