leetcode Different Ways to Add Parentheses

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)