Different Ways to Add Parentheses-优快云博客

本文介绍了一种使用递归算法解决LeetCode题目中给定包含数字与运算符的字符串，计算所有可能组合结果的方法。通过拆分字符串并分别计算左右两边表达式的不同结果来实现。

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题目来源LeetCode

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]

这道题是递归的方法从前往后遍历，遇到运算符就将其左右两边子字符串分成两部分，然后计算两边的不同结果，代码如下：

class Solution {  
public:  
   vector<int> diffWaysToCompute(string input) {  
       vector<int>result;  
       int len = input.length();
       for(int i = 0; i < len; i++) { 
           if(input[i] < '0' || input[i] > '9') {  
          vector<int>lp = diffWaysToCompute(input.substr(0,i));  
          vector<int>rp = diffWaysToCompute(input.substr(i+1)); 
               for(int j = 0 ; j < lp.size(); j++) { 
               for(int q = 0; q < rp.size(); q++) { 
                   if(input[i] == '+') {  
                       result.push_back(lp[j] + rp[q]);  
                    } 
                   else if(input[i] == '-') { 
                       result.push_back(lp[j] - rp[q]); 
                   }
                   else {  
                       result.push_back(lp[j] * rp[q]);  
                   }  
                }  
                }  
        }  
        }  
        if(result.empty())
        result.push_back(atoi(input.c_str()));  
        return result;  
    }  
};