241. Different Ways to Add Parentheses

Question

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"
Output: [0, 2]
Explanation:
((2-1)-1) = 0
(2-(1-1)) = 2

Example 2:

Input: "2*3-4*5"
Output: [-34, -14, -10, -10, 10]
Explanation:
(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

Thinking:

  • Method 1:分治
class Solution {
    public List<Integer> diffWaysToCompute(String input) {
        List<Integer> res = new ArrayList<>();
        int len = input.length();
        for(int i = 0; i < len; i++){
            char c = input.charAt(i);
            if(c == '*' || c == '-' || c == '+'){
                List<Integer> left = diffWaysToCompute(input.substring(0, i));
                List<Integer> right = diffWaysToCompute(input.substring(i + 1, len));
                for(Integer l : left){
                    for(Integer r : right){
                        if(c == '*') res.add(l * r);
                        else if(c == '+') res.add(l + r);
                        else if(c == '-') res.add(l - r);
                    }
                }
            }
        }
        if(res.size() == 0) res.add(Integer.parseInt(input));
        return res;
    }
}

Second time

  1. Still need to use devide and conquer method, I review previous answer.
class Solution {
    public List<Integer> diffWaysToCompute(String input) {
        List<Integer> result = new LinkedList<>();
        int len = input.length();
        for(int i = 0; i < len; i++){
            char c = input.charAt(i);
            if(c == '+' || c == '-' || c == '*'){ //Devide at operators.
                List<Integer> lefts = diffWaysToCompute(input.substring(0, i)); //If possible, divide to a small question.
                List<Integer> rights = diffWaysToCompute(input.substring(i + 1, len));
                for(Integer l : lefts){ //Merge, all left values can match all right values, so we use the operators to merge the results.
                    for(Integer r: rights){
                        if(c == '+') result.add(l + r);
                        else if(c == '-') result.add(l - r);
                        else result.add(l * r);
                    }
                }
            }
        }
        if(result.isEmpty()) result.add(Integer.parseInt(input)); //if their is nothing to devide, it means it is a single number, we can add the number into the list, which is the minimum sub-question to solve.(Conquer)
        return result;
    }
}