Different Ways to Add Parentheses

Given a string of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. The valid operators are+, - and *.

Example 1

Input: "2-1-1".

((2-1)-1) = 0(2-(1-1)) = 2

Output: [0, 2]

Example 2

Input: "2*3-4*5"

(2*(3-(4*5))) = -34((2*3)-(4*5)) = -14((2*(3-4))*5) = -10(2*((3-4)*5)) = -10(((2*3)-4)*5) = 10

Output: [-34, -14, -10, -10, 10]

这个题不难，将操作符的两边的结果计算出来，然后再用操作符运算即可。

如果input中不包含操作符，说明只有数字，输出结果。

public class Solution {    public List<Integer> diffWaysToCompute(String input){List<Integer> list=new ArrayList<Integer>();boolean isDigit=true;for(int i=0;i<input.length();i++){if(input.charAt(i)=='+'||input.charAt(i)=='-'||input.charAt(i)=='*'){isDigit=false;}}if(isDigit){list.add(Integer.parseInt(input));return list;}for(int i=1;i<input.length()-1;i++){List<Integer> leftList=diffWaysToCompute(input.substring(0, i));List<Integer> rightList=diffWaysToCompute(input.substring(i+1));for(int j=0;j<leftList.size();j++){for(int k=0;k<rightList.size();k++){if(input.charAt(i)=='+')list.add(leftList.get(j)+rightList.get(k));if(input.charAt(i)=='-')list.add(leftList.get(j)-rightList.get(k));if(input.charAt(i)=='*')list.add(leftList.get(j)*rightList.get(k));}}}return list;}}