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]
解决方法：采取分治的方法，根据’+‘’-‘’*‘将字符串分为左右两部分，然后再进行合并，采取递归的方法进行编程，当最后分治只剩下一个数字的时候，此时list的size为0，则认为这是这个最后分支的最终结果。具体JAVA代码如下：

public  List<Integer> diffWaysToCompute(String input) {        List<Integer>ret=new ArrayList<Integer>();        for (int i = 0; i < input.length(); i++) {            if (input.charAt(i)=='+'||input.charAt(i)=='-'||input.charAt(i)=='*') {                String part1=input.substring(0,i);                String part2=input.substring(i+1);                List<Integer> list1= diffWaysToCompute(part1);                List<Integer> list2= diffWaysToCompute(part2);                for (Integer p1:list1) {                    for (Integer p2:list2) {                        int c=0;                        switch (input.charAt(i)) {                        case '+':                            c=p1+p2;                                                        break;                        case '-':                            c=p1-p2;                            break;                        case '*':                            c=p1*p2;                            break;                                              }                        ret.add(c);                    }                }            }        }        if (ret.size()==0) {            ret.add(Integer.valueOf(input));        }                      return ret;    }