Different Ways to Add Parentheses

题目地址：
https://leetcode.com/problems/different-ways-to-add-parentheses/description/

题目：

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]

解题思路：
这道题就是用divide and conquer的解法，然后可以用map来做memorization。

代码：

public class DifferentWaystoAddParentheses {

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

    public List<Integer> diffWaysToCompute(String input) {
        if(map.containsKey(input)){
            return map.get(input);
        }
        List<Integer> rst = new ArrayList<Integer>();
        for(int i = 0; i < input.length(); i++){
            char c = input.charAt(i);
            if(c == '+' || c == '-' || c == '*'){
                List<Integer> list1 = diffWaysToCompute(input.substring(0, i));
                List<Integer> list2 = diffWaysToCompute(input.substring(i + 1));
                for(int v1 : list1){
                    for(int v2 : list2){
                        if(c == '+'){
                            rst.add(v1 + v2);
                        }
                        if(c == '-'){
                            rst.add(v1 - v2);
                        }
                        if(c == '*'){
                            rst.add(v1 * v2);
                        }
                    }
                }
            }
        }

        if(rst.isEmpty()){
            rst.add(Integer.parseInt(input));
        }

        map.put(input, rst);
        return rst;
    }
}