241. Different Ways to Add Parentheses

Total Accepted: 23674  Total Submissions: 65569  Difficulty: Medium

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]

Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.

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分析：

本题不会，代码纯属借鉴和学习！

这是一个要考虑各种组合情况的问题，关键还是怎么入手得到各种组合情况！

从别人的代码来看，一旦遇到运算符号(不是运算符号暂时不处理)，就分治法考虑各种组合情况：

然后将运算结果重新根据当前运算符进行运算。

class Solution {
public:
    vector<int> diffWaysToCompute(string input) {
        vector<int> result;
        int size = input.size();
        for (int i = 0; i < size; i++) {
            char cur = input[i];
            if (cur == '+' || cur == '-' || cur == '*') {//如果是运算符号
                // 分治法，分两边解决，解决获取结果之后重新根据当前符号进行运算
                vector<int> result1 = diffWaysToCompute(input.substr(0, i));//从0位置开始获取长度为i的字符串
                vector<int> result2 = diffWaysToCompute(input.substr(i+1));//从0位置开始获取后面所有字符
                for (auto n1 : result1) {
                    for (auto n2 : result2) {
                        if (cur == '+')
                            result.push_back(n1 + n2);
                        else if (cur == '-')
                            result.push_back(n1 - n2);
                        else
                            result.push_back(n1 * n2);    
                    }
                }
            }
        }
        // 如果输入字符串只是数字（字符数字）
        if (result.empty())
            result.push_back(atoi(input.c_str()));
        return result;
    }
};

完整代码：

#include<iostream>
#include<vector>
using namespace std;

vector<int> diffWaysToCompute(string input) {
    vector<int> result;
    int len=input.size();
    for(int k=0;k<len;k++)
    {
        if(input[k]=='+'||input[k]=='-'||input[k]=='*')
        {
            vector<int> result1=diffWaysToCompute(input.substr(0,k));
            vector<int> result2=diffWaysToCompute(input.substr(k+1));
            for(vector<int>::iterator i=result1.begin();i!=result1.end();i++)
                for(vector<int>::iterator j=result2.begin();j!=result2.end();j++)
                {
                    if(input[k]=='+')
                        result.push_back((*i)+(*j));
                    else if(input[k]=='-')
                        result.push_back((*i)-(*j));
                    else
                        result.push_back((*i)*(*j));
                }
        }
    }
    if(result.empty())
        result.push_back(atoi(input.c_str()));
    return result;
}
int main()
{
    string input="2*3-4*5";
    vector<int> vec;
    vec=diffWaysToCompute(input);
    for(int i=0;i<vec.size();i++)
        cout<<vec[i]<<' ';
    cout<<endl;
    system("pause");
}

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原文地址：http://blog.csdn.net/ebowtang/article/details/51581228

原作者博客：http://blog.csdn.net/ebowtang

本博客LeetCode题解索引：http://blog.csdn.net/ebowtang/article/details/50668895