LeetCode: 241. Different Ways to Add Parentheses

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题目描述

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:

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

Example 2:

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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

这题需要使用分治法。比如对于2*3-4*5，可以根据操作符来进行分治，比如根据’-‘可以分成2*3和4*5，然后再算每一个小块的所有可能值，最后用’-‘在所有可能值之间进行减法操作。

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class Solution {
private:
    vector<int> getInt(string input) {
        vector<int> result;
        int startIndex = 0;

        for (int i = 0; i < input.length(); i++) {
            if (input[i] == '+' || input[i] == '-' || input[i] == '*'){
                result.push_back(stoi(input.substr(startIndex, i-startIndex)));
                startIndex = i+1;
            }
        }
        result.push_back(stoi(input.substr(startIndex, input.length()-startIndex)));
        return result;
    }

    vector<char> getChar(string input) {
        vector<char> result;
        for (int i = 0; i < input.length(); i++) {
            if (input[i] == '+' || input[i] == '-' || input[i] == '*')
                result.push_back(input[i]);
        }
        return result;
    }

    vector<int> divide(vector<int> ints, vector<char> chars, int left, int right) {
        vector<int> result;

        if (left == right) {
            result.push_back(ints[left]);

            return result;
        }

        for (int i=left; i<right; i++) {
            vector<int> leftVector = divide(ints, chars, left, i);
            vector<int> rightVector = divide(ints, chars, i+1, right);

            for (int j: leftVector) {
                for (int k: rightVector) {
                    if (chars[i] == '+')
                        result.push_back(j+k);
                    if (chars[i] == '-')
                        result.push_back(j-k);
                    if (chars[i] == '*')
                        result.push_back(j*k);
                }
            }
        }

        return result;
    }

public:
    vector<int> diffWaysToCompute(string input) {
        vector<int> ints = getInt(input);
        vector<char> chars = getChar(input);

        return divide(ints, chars, 0, ints.size()-1);
    }
};