[LeetCode] 241. Different Ways to Add Parentheses 添加括号的不同方式

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]

给定一个含有数字和运算符的表达式，运算符可以是加减乘，在任意位置添加括号，求出所有可能的表达式值。 

解法：递归

Java:

public class Solution {
    public List<Integer> diffWaysToCompute(String input) {
        List<Integer> result = new ArrayList<>();
        if (input == null || input.length() == 0) {
            return result;
        }
         
        for (int i = 0; i < input.length(); i++) {
            char c = input.charAt(i);
             
            if (!isOperator(c)) {
                continue;
            }
             
            List<Integer> left = diffWaysToCompute(input.substring(0, i));
            List<Integer> right = diffWaysToCompute(input.substring(i + 1));
             
            for (int num1 : left) {
                for (int num2 : right) {
                    int val = calculate(num1, num2, c);
                    result.add(val);
                }
            }
        }
         
        // only contains one number
        if (result.isEmpty()) {
            result.add(Integer.parseInt(input));
        }
         
        return result;
    }
     
    private int calculate(int num1, int num2, char operator) {
        int result = 0;
         
        switch(operator) {
            case '+' : result = num1 + num2;
            break;
             
            case '-' : result = num1 - num2;
            break;
             
            case '*' : result = num1 * num2;
            break;
        }
         
        return result;
    }
     
    private boolean isOperator(char operator) {
        return (operator == '+') || (operator == '-') || (operator == '*');
    }
}　　

Python:

import operator
import re

class Solution:
    # @param {string} input
    # @return {integer[]}
    def diffWaysToCompute(self, input):
        tokens = re.split('(\D)', input)
        nums = map(int, tokens[::2])
        ops = map({'+': operator.add, '-': operator.sub, '*': operator.mul}.get, tokens[1::2])
        lookup = [[None for _ in xrange(len(nums))] for _ in xrange(len(nums))]

        def diffWaysToComputeRecu(left, right):
            if left == right:
                return [nums[left]]
            if lookup[left][right]:
                return lookup[left][right]
            lookup[left][right] = [ops[i](x, y)
                                   for i in xrange(left, right)
                                   for x in diffWaysToComputeRecu(left, i)
                                   for y in diffWaysToComputeRecu(i + 1, right)]
            return lookup[left][right]

        return diffWaysToComputeRecu(0, len(nums) - 1)　　

Python:

class Solution:
    # @param {string} input
    # @return {integer[]}
    def diffWaysToCompute(self, input):
        lookup = [[None for _ in xrange(len(input) + 1)] for _ in xrange(len(input) + 1)]
        ops = {'+': operator.add, '-': operator.sub, '*': operator.mul}

        def diffWaysToComputeRecu(left, right):
            if lookup[left][right]:
                return lookup[left][right]
            result = []
            for i in xrange(left, right):
                if input[i] in ops:
                    for x in diffWaysToComputeRecu(left, i):
                        for y in diffWaysToComputeRecu(i + 1, right):
                            result.append(ops[input[i]](x, y))

            if not result:
                result = [int(input[left:right])]
            lookup[left][right] = result
            return lookup[left][right]

        return diffWaysToComputeRecu(0, len(input))　

C++:

class Solution {
public:
    vector<int> diffWaysToCompute(string input) {
        vector<int> res;
        for (int i = 0; i < input.size(); ++i) {
            if (input[i] == '+' || input[i] == '-' || input[i] == '*') {
                vector<int> left = diffWaysToCompute(input.substr(0, i));
                vector<int> right = diffWaysToCompute(input.substr(i + 1));
                for (int j = 0; j < left.size(); ++j) {
                    for (int k = 0; k < right.size(); ++k) {
                        if (input[i] == '+') res.push_back(left[j] + right[k]);
                        else if (input[i] == '-') res.push_back(left[j] - right[k]);
                        else res.push_back(left[j] * right[k]);
                    }
                }
            }
        }
        if (res.empty()) res.push_back(atoi(input.c_str()));
        return res;
    }
};

　　

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