241. 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]

递归算法：
遇到一个运算符就将其两侧分开，进行递归运算返回结果为这两侧进行运算后的各种结果的list

遍历这个list里的值 进行运算 

public class Solution {
public List<Integer> diffWaysToCompute(String input) {
List<Integer> list=new ArrayList<Integer>();
for(int i=0;i<input.length();i++){
char a=input.charAt(i);
if(a=='+'||a=='-'||a=='*'){
String s1=input.substring(0,i);
String s2=input.substring(i+1);
List<Integer> a1=diffWaysToCompute(s1);    //返回的结果是 运算结果集合 不管过程 用它的值进行计算
List<Integer> a2=diffWaysToCompute(s2);
for(int m:a1){
for(int n:a2){
if(a=='+')
list.add(m+n);
else if(a=='-')
list.add(m-n);
else if(a=='*')
list.add(m*n);
}
}
}
}
if(list.size()==0) list.add(Integer.valueOf(input));
return list;
}
}