一、题目

Given a string expression of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. You may return the answer in any order.Example 1:Input: expression = "2-1-1"Output: [0,2]Explanation:
((2-1)-1) = 0
(2-(1-1)) = 2Example 2:Input: expression = "2*3-4*5"Output: [-34,-14,-10,-10,10]Explanation:
(2*(3-(4*5))) = -34
((2*3)-(4*5)) = -14
((2*(3-4))*5) = -10
(2*((3-4)*5)) = -10
(((2*3)-4)*5) = 10Constraints:1 <= expression.length <= 20expression consists of digits and the operator '+', '-', and '*'.All the integer values in the input expression are in the range [0, 99].

二、思路

首先题目叫求添加括号来计算结果，可以有重复值，需要将所有可以添加括号的可能遍历
可以分解为子问题，遇到一个符号+,−,∗+,-,*+,−,∗那么计算左边字符串可能的取值和右边字符串可能的取值，然后合并得出当前字符串可能的取值
对于一个字符串需要遍历所有的运算符号来获得所有的情况
递归出口：当一个字符串没有符号时即全部为数字，那么只有一种取值即字符串本身

三、代码

class Solution {public:vector<int> diffWaysToCompute(string expression) {vector<int> ans;for (int i = 0; i < expression.size(); i++) {if (!(expression[i] >= '0' && expression[i] <= '9')) {//切割为左右vector<int> l = diffWaysToCompute(expression.substr(0, i));vector<int> r = diffWaysToCompute(expression.substr(i + 1, expression.size()));//将左右的结果保存在ansfor (int j = 0; j < l.size(); j++) {for (int k = 0; k < r.size(); k++) {switch(expression[i]) {case '+':ans.push_back(l[j] + r[k]);break;case '-':ans.push_back(l[j] - r[k]);break;case '*':ans.push_back(l[j] * r[k]);break;}}}}}   if (ans.size() == 0) {//递归出口代表只有一个数字 这时候直接将数字进行转化ans.push_back(stoi(expression));}return ans;}
};

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LeedCode 241. Different Ways to Add Parentheses

一、题目

二、思路

三、代码

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