241. Different Ways to Add Parentheses

最新推荐文章于 2021-11-19 22:29:55 发布

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本文介绍了一种使用递归算法解决给定包含数字和运算符的字符串，计算所有可能组合结果的方法。通过将输入字符串转换为数字和运算符列表，并运用备忘录技术减少重复计算，最终返回所有可能的计算结果。

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]

Solution:

It is hard to solve it iteratively, using recursion would be simpler, for each of the operation just compute all the possible answers on the left and right and combine them together.

Code:

public class Solution {
    public List<Integer> diffWaysToCompute(String input) {
        List<String> strs = convert(input);
        HashMap<Integer,List<Integer>> hm = new HashMap();
        return helper(strs,0,strs.size() - 1,hm);
       
    }
    
    public List<Integer> helper(List<String> strs, int from, int to, HashMap<Integer,List<Integer>> hm){
        int key = from * strs.size() + to;
        if(hm.containsKey(key)){
            return hm.get(key);
        }
        
        List<Integer> l = new ArrayList<Integer>();
        
        if(from == to){
            l.add(Integer.parseInt(strs.get(from)));
            hm.put(key,l);
            return l;
        }
        
        for(int i = from + 1; i < to; i+=2){
            List<Integer> i1 = helper(strs,from,i-1,hm);
            List<Integer> i2 = helper(strs,i + 1, to,hm);
            if(strs.get(i).equals("+")){
                for(int j1 : i1){
                    for(int j2 : i2){
                        l.add(j1 + j2);
                    }
                }
                
            } else if(strs.get(i).equals("-")){
                for(int j1 : i1){
                    for(int j2 : i2){
                        l.add(j1 - j2);
                    }
                }
            } else {
                for(int j1 : i1){
                    for(int j2 : i2){
                        l.add(j1 * j2);
                    }
                }
            }
        }
        
        hm.put(key,l);
        return l;
        
    }
    
    
    
    public List<String> convert(String s){
        List<String> ret = new ArrayList<>();
        int num = 0;
        for(char c : s.toCharArray()){
            if(Character.isDigit(c)){
                num = num * 10 + c - '0';
            } else {
                ret.add("" + num);
                ret.add("" + c);
                num = 0;
            }
        }
        ret.add("" + num);
        return ret;
    }
    
}