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]);}elsereturn 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]);elseresult.push_back(leftRes[j]*rightRes[k]);}}}dp[left][right]=result;return result;}};

他山之石：

;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]);elseret.push_back(left[j] * right[k]);}}}}if (ret.empty())ret.push_back(atoi(input.c_str()));return ret;}};

，爬上那座山，听最圣洁的经。