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"
Output: [0, 2]
Explanation:
((2-1)-1) = 0
(2-(1-1)) = 2
Example 2:
Input:"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) = 10
题目大意:
对一个含有运算符'+''-''*'的运算表达式,通过添加‘( )’改变运算符优先级,计算所有可能的结果(包括重复的结果)。
理解:
采用递归和分而治之的方法,遍历整个字符串表达式,每遇到一个运算符,将它左边表达式作为一个,右边表达式作为另一个整体,依次递归计算。
函数返回值为当前表达式所有可能的计算结果,左部OPERATOR右部,即为所有可能结果值。
代码C++:
class Solution { public: vector<int> diffWaysToCompute(string input) { vector<int> res; int len = input.length(); for(int i=0;i<len;++i){ if(input[i]=='+' ||input[i]=='-' || input[i]=='*'){ auto res1 = diffWaysToCompute(input.substr(0,i)); auto res2 = diffWaysToCompute(input.substr(i+1)); for(auto left:res1){ for(auto right:res2){ switch(input[i]){ case '+': res.push_back(left+right); break; case '-': res.push_back(left-right); break; case '*': res.push_back(left*right); break; } } } } } if(res.empty()){ res.push_back(stoi(input)); } return res; } };