Description：

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]

Credits:

Special thanks to  for adding this problem and creating all test cases.

使用分治递归的方法首先将问题划分成子问题，再对子问题的解进行全排列合并，最终得到问题的解。

深刻理解递归还是很有必要的。

public class Solution {    public List
     
       diffWaysToCompute(String input) {                List
      
        resList = new ArrayList
       
        ();                                for(int i=0; i
        
          leftList = diffWaysToCompute(input.substring(0, i));                List
         
           rightList = diffWaysToCompute(input.substring(i+1));                                for(int left : leftList) {                    for(int right : rightList) {                        if(op == '+') {                            resList.add(left + right);                         }                        else if(op == '-') {                            resList.add(left - right);                        }                        else {                            resList.add(left * right);                        }                    }                }                                            }        }                if(resList.size() == 0) {            resList.add(Integer.parseInt(input));        }                return resList;    }}
         
        
       
      
     

这题竟然没有不是个位数的case。