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)