leetcode Different Ways to Add Parentheses

xiaoxiao2021-02-28  3

Different Ways to Add Parentheses

题目详情:

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]

解题方法:

典型的分治算法,当遇到运算符时分为左右两边进行相同的运算。之后再在将左右两边的数值分别两两进行运算。

代码详情:

class Solution { public:     vector<int> diffWaysToCompute(string input) {         vector<int> result;         int size = input.size();         for (int i = 0; i < size; i++) {             char cur = input[i];             if (cur == '+' || cur == '-' || cur == '*') {                 // Split input string into two parts and solve them recursively                 vector<int> result1 = diffWaysToCompute(input.substr(0, i));                 vector<int> result2 = diffWaysToCompute(input.substr(i+1));                 for (int p = 0; p < result1.size(); p++) {                     for (int q = 0; q < result2.size(); q++) {                         if (cur == '+')                             result.push_back(result1[p]+result2[q]);                         else if (cur == '-')                             result.push_back(result1[p]-result2[q]);                         else                             result.push_back(result1[p]*result2[q]);                         }                 }             }         }         // if the input string contains only number         if (result.empty())             result.push_back(atoi(input.c_str()));         return result;     }      };

复杂度分析:

设有n个num,计算x个num的时间为f(x),则复杂度为[f(1)+f(n-1)+f(2)+f(n-2)+f(3)+f(n-3)+.....+f(n/2)+f(n/2)]*2 = f(n)

转载请注明原文地址: https://www.6miu.com/read-2000318.html

最新回复(0)