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; }