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;
}