[leetcode] 241. Different Ways to Add Parentheses 解题报告

题目链接: https://leetcode.com/problems/different-ways-to-add-parentheses/

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]




思路: 使用分治法, 以每个符号为分割, 将其分为左右两部分, 每一部分返回一个数组, 代表这一部分的运算结果有几种, 然后再用这个分割的符号, 将左右两边的各种结果以这种符号运算, 保存到一个新的数组中, 然后返回就是了. 为了简便起见, 我先将字符串拆分成了字符串数组, 然后再进行分治运算, 我看到有的人直接是边分割边计算的, 那样当然更好.

代码如下:

class Solution {
public:
    vector<int> divid(int left, int right)
    {
        vector<int> result;
        if(left==right)
        {
            result.push_back(stoi(str[left]));
            return result;
        }
        for(int i = left+1; i<= right-1; i+=2)
        {
            vector<int> tem1 = divid(left, i-1), tem2 = divid(i+1,right);
            for(auto val1: tem1)
            {
                for(auto val2: tem2)
                {
                    if(str[i]=="+") result.push_back(val1 + val2);
                    else if(str[i]=="-") result.push_back(val1 - val2);
                    else if(str[i]=="*") result.push_back(val1 * val2);
                }
            }
        }
        
        return result;
    }
    vector<int> diffWaysToCompute(string input) {
        if(input.size()==0) return vector<int>();
        int st=0, i =-1;
        while(++i < input.size())
        {
            if(input[i]!='+' && input[i]!='-' && input[i]!='*') continue;
            string tem = input.substr(st, i-st);
            st = i+1;
            str.push_back(tem);
            str.push_back(input.substr(i, 1));
        }
        str.push_back(input.substr(st));
        return divid(0, str.size()-1);
    }
    
private:
    vector<string> str;
};