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 *
.
Input: "2-1-1"
.
((2-1)-1) = 0 (2-(1-1)) = 2
Output: [0, 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]
Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.
解题思路;
这道题可以用递归的方法来做。对于一个输入字符串s,一次获得每个标点符号的左侧left和右侧right的值,然后两两组合成结果。
class Solution { public: vector<int> diffWaysToCompute(string input) { vector<int> result; int len = input.length(); if(len == 0){ return result; } if(isInt(input)){ result.push_back(std::stoi(input)); return result; } for(int i=0; i<len; i++){ if(input[i]<'0' || input[i]>'9'){ vector<int> leftResult = diffWaysToCompute(input.substr(0, i)); vector<int> rightResult = diffWaysToCompute(input.substr(i+1)); for(int m = 0; m<leftResult.size(); m++){ for(int n = 0; n<rightResult.size(); n++){ switch(input[i]){ case '+': result.push_back(leftResult[m] + rightResult[n]); break; case '-': result.push_back(leftResult[m] - rightResult[n]); break; case '*': result.push_back(leftResult[m] * rightResult[n]); break; } } } } } return result; } bool isInt(string& s){ int len = s.length(); for(int i=0; i<len; i++){ if(s[i]<'0' || s[i]>'9'){ return false; } } return true; } };
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[LeetCode] Different Ways to Add Parentheses
原文地址:http://blog.csdn.net/kangrydotnet/article/details/47722565