Leetcode 241- 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]

class Solution {
public:
    vector<int> diffWaysToCompute(string input) {
        vector<int> result;
        vector<int> nums;
        vector<char> operate;
        int pre=0;
        for(int i=0;i<input.length();i++)
        {
            if(i==0&&input[0]<‘0‘||input[0]>‘9‘)
                return result;
            if(input[i]>=‘0‘&&input[i]<=‘9‘)
            {
                pre=pre*10+input[i]-‘0‘;
                if(i==input.length()-1)
                    nums.push_back(pre);
            }
            else if(input[i]==‘+‘||input[i]==‘-‘||input[i]==‘*‘)
            {
                nums.push_back(pre);
                pre=0;
                operate.push_back(input[i]);
            }
            else
                return result;
        }
        int count=nums.size();
        vector<vector<vector<int>>>dp(count,vector<vector<int>>(count));

        result=compute(nums, operate, dp, 0, count-1);
        return result;
    }

    vector<int> compute(vector<int> &nums, vector<char> &operate, vector<vector<vector<int>>> &dp, int left, int right)
    {
        vector<int> result;
        if(left==right)
        {
            result.push_back(nums[left]);
            return result;
        }

        for(int i=left+1;i<=right;i++)
        {
            vector<int> leftRes,rightRes;
            leftRes=(!dp[left][i-1].empty())?dp[left][i-1]:compute(nums,operate,dp,left,i-1);
            rightRes=(!dp[i][right].empty())?dp[i][right]:compute(nums,operate,dp,i,right);
            for(int j=0;j<leftRes.size();j++)
            {
                for(int k=0;k<rightRes.size();k++)
                {
                    if(operate[i-1]==‘+‘)
                        result.push_back(leftRes[j]+rightRes[k]);
                    else if(operate[i-1]==‘-‘)
                        result.push_back(leftRes[j]-rightRes[k]);
                    else
                        result.push_back(leftRes[j]*rightRes[k]);
                }
            }
        }
        dp[left][right]=result;
        return result;
    }
};

他山之石：

#include <vector>
using namespace std;
class Solution {
public:
    vector<int> diffWaysToCompute(string input) {
        vector<int> ret;
        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] == ‘+‘)
                            ret.push_back(left[j] + right[k]);
                        else if (input[i] == ‘-‘)
                            ret.push_back(left[j] - right[k]);
                        else
                            ret.push_back(left[j] * right[k]);
                    }
                }
            }
        }
        if (ret.empty())
            ret.push_back(atoi(input.c_str()));
        return ret;
    }
};