leetcode241 : Different Ways to Add Parentheses

1、原题如下：
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]

2、解题如下：

class Solution {public:    vector<int> diffWaysToCompute(string input) {        vector<int> result;        int offset=0;//用来弥补string中没有+-*的情况        for(int i=0;i<input.size();i++)        {            if(input[i]=='+'||input[i]=='-'||input[i]=='*')            {                offset=1;                vector<int> temp1=diffWaysToCompute(input.substr(0,i));                vector<int> temp2=diffWaysToCompute(input.substr(i+1));                for(auto x:temp1)                {                    for(auto y:temp2)                    {                        if(input[i]=='+')                            result.push_back(x+y);                        if(input[i]=='-')                            result.push_back(x-y);                        if(input[i]=='*')                            result.push_back(x*y);                    }                }            }        }        if(offset==0)        {            result.push_back(stoi(input));        }        return result;    }};

3、总结
循环迭代的算法题目很多，做到这会儿已经更加轻车熟路了，只需要注意边界条件即可，这里采用了offset来讨论边界问题