[LeetCode]241.Different Ways to Add Parenthese
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description
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]解题思路
这道题要求解出输入多项式以所有的方式组合数字和计算符得到的解。这里运用分治法,通过迭代,可以得到答案。对于一个多项式,总有一个运算符是在最后进行运算的,当两侧的算式都得到答案后,再根据该运算符进行计算得到最终答案。那么同样,两侧的算式中每个算式也有一个运算符是最后一个运算的,当该运算符两侧的算式得到答案后,进行该计算得到该侧的答案。我们想要得到所有的组合方式的答案,那么我们首先需要遍历一遍整个多项式,将每个运算符都作为最后计算的运算符,分别计算出该运算符两侧的结果,这里通过迭代得到两侧的算式以不同方式组合得到的结果,将两侧的结果进行最后的计算,得到最终答案。代码如下
class Solution {public: vector<int> diffWaysToCompute(string input) { vector<int> answer; int n = input.size(); for(int i = 0; i < n; i++){ if(input[i] == '+' || input[i] == '-' || input[i] =='*'){ //得到左侧式子以不同方式计算的结果的向量 vector<int> left = diffWaysToCompute(input.substr(0, i)); //得到右侧式子以不同方式计算的结果的向量 vector<int> right = diffWaysToCompute(input.substr(i+1)); //将左右两侧的结果分别进行计算 for(int m = 0; m < left.size();m++){ for(int n = 0; n < right.size(); n++){ if(input[i] == '+') answer.push_back(left[m]+right[n]); else if(input[i] == '-') answer.push_back(left[m]-right[n]); else answer.push_back(left[m]*right[n]); } } } } if(answer.size() == 0){ answer.push_back(atoi(input.c_str())); } return answer; }};
LeetCode题目原址
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