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]

Credits:

Special thanks to @mithmatt for adding this problem and creating all test cases.
分治法解题。每个符号，分成左边和右边，左边所有可能的值 + - * 右边左右的值，得到所有可能的答案。代码如下：
public class Solution {

HashMap<String, List<Integer>> map = new HashMap<String, List<Integer>>();

public List<Integer> diffWaysToCompute(String input) {
List<Integer> res = new ArrayList<Integer>();
for (int i = 0; i < input.length(); i ++) {
char ch = input.charAt(i);
if (ch == '+' || ch == '-' || ch == '*') {
String leftString = input.substring(0, i);
String rightString = input.substring(i + 1);
List<Integer> l = map.getOrDefault(leftString, diffWaysToCompute(leftString));
List<Integer> r = map.getOrDefault(rightString, diffWaysToCompute(rightString));
for (int lnum: l) {
for (int rnum: r) {
if (ch == '+') {
res.add(lnum + rnum);
} else if (ch == '-') {
res.add(lnum - rnum);
} else if (ch == '*') {
res.add(lnum * rnum);
}
}
}
}
}
if (res.size() == 0) {
res.add(Integer.valueOf(input));
}
map.put(input, res);
return res;
}
}