leetcode241. Different Ways to Add Parentheses

题目要求

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]

如今有一个字符串形式的算式，求该算式在任意位置加上任意数量的括号后可以得出的全部可能的值。

思路与算法

这是经典的分治法。咱们将算式从一个操做符的位置分割开，并分别得出左边和右边算式全部可能的值。再将左右的值分别按照操做符进行计算。这里经过递归实现。

public List<Integer> diffWaysToCompute(String input) {
        return diffWaysToCompute(input, 0, input.length());
    }
    
    public List<Integer> diffWaysToCompute(String input, int startIndex, int endIndex){
        boolean isDigit = true;
        List<Integer> result = new ArrayList<Integer>();
        for(int i = startIndex ; i<endIndex ; i++){
            char cur = input.charAt(i);
            if(cur == '+' || cur == '-' || cur=='*' ){
                isDigit = false;
                List<Integer> leftValue = diffWaysToCompute(input, startIndex, i);
                List<Integer> rightValue = diffWaysToCompute(input, i+1, endIndex);
                result.addAll(compute(leftValue, rightValue,cur));
            }
        }
        if(isDigit){
            result.add(Integer.parseInt(input.substring(startIndex, endIndex)));
        }
        return result;
    }
    
    public List<Integer> compute(List<Integer> leftValue, List<Integer> rightValue, char operator){
        switch(operator){
        case '+' : return add(leftValue, rightValue);
        case '-' : return minus(leftValue, rightValue);
        case '*' : return multiply(leftValue, rightValue);
        }
        return new ArrayList<>();
    }
    
    public List<Integer> add(List<Integer> leftValue, List<Integer> rightValue){
        List<Integer> result = new ArrayList<Integer>();
        for(int left : leftValue){
            for(int right : rightValue){
                result.add(left + right);
            }
        }
        return result;
    }
    
    public List<Integer> minus(List<Integer> leftValue, List<Integer> rightValue){
        List<Integer> result = new ArrayList<Integer>();
        for(int left : leftValue){
            for(int right : rightValue){
                result.add(left - right);
            }
        }
        return result;
    }
    
    public List<Integer> multiply(List<Integer> leftValue, List<Integer> rightValue){
        List<Integer> result = new ArrayList<Integer>();
        for(int left : leftValue){
            for(int right : rightValue){
                result.add(left * right);
            }
        }
        return result;
    }

提升性能的方法是经过Map实现缓存。将已经遍历过的结果存入缓存中，若是碰到重复的计算式，就能够直接获取其全部可能的值。

Map<String, List<Integer>> cache = new HashMap<String, List<Integer>>();
    public List<Integer> diffWaysToCompute_withCache(String input){
        if(cache.containsKey(input)) return cache.get(input);
        List<Integer> result = new ArrayList<Integer>();
        boolean isDigit = true;
        for(int i = 0 ; i<input.length() ; i++){
            int cur = input.charAt(i);
            if(cur == '+' || cur == '-' || cur == '*'){
                isDigit = false;
                List<Integer> leftValues = diffWaysToCompute_withCache(input.substring(0, i));
                List<Integer> rightValues = diffWaysToCompute_withCache(input.substring(i+1));
                for(Integer left : leftValues){
                    for(Integer right : rightValues){
                        switch(cur){
                        case '+' : result.add(left + right); break;
                        case '-' : result.add(left - right); break;
                        case '*' : result.add(left * right); break;
                        }
                    }
                }
            }
        }
        if(isDigit){ result.add(Integer.parseInt(input));}
        cache.put(input, result);
        return result;
    }

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