问题：
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) = 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:
- 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.
解决：
【题意】 给定一段连续的除法式子，我们需要给它在不同地方加括号，确保能得到最大的结果，最后以string的形式返回结果，要求不能存在冗余括号。
① 除了第一个除数之外，后面的数都可以转变为乘积！！！
拿样例来说：
1000/(100/10/2) == (1000*10*2)/(100) 所以，我们只需要考虑三种情况： 1. 只有一个数，直接返回； 2. 有两个数，第一个除以第二个返回； 3. 有三个及以上的数，把第二个数后面的和第一个数全部乘起来，最后除以第二个数。（因为note当中说明了，给的数字都是[2,1000]的，所以第二个数后面的所有数乘起来都只会让结果变大）。 class Solution { //9ms
public String optimalDivision(int[] nums) { String res = ""; res = String.valueOf(nums[0]); if (nums.length == 1) return res; if (nums.length == 2) return res + "/" + String.valueOf(nums[1]); res += "/(" + String.valueOf(nums[1]); for (int i = 2;i < nums.length;i ++){ res += "/" + String.valueOf(nums[i]); } res += ")"; return res; } }② 使用StringBuilder类保存结果。
public class Solution { //9ms
public String optimalDivision(int[] nums) { if (nums.length == 1) return nums[0] + ""; if (nums.length == 2) return nums[0] + "/" + nums[1]; StringBuilder res = new StringBuilder(nums[0] + "/(" + nums[1]); for (int i = 2; i < nums.length; i++) { res.append("/" + nums[i]); } res.append(")"); return res.toString(); } }