标签：input public tin operator nbsp str 情况 push span
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"Output:[0, 2]Explanation: ((2-1)-1) = 0 (2-(1-1)) = 2
Example 2:
Input:"2*3-4*5"Output:[-34, -14, -10, -10, 10]Explanation: (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
长见识了，大概知道分治的用法了， 额，求二叉树的高度的思维和 分治很像啊， 快排应该就是分治算法的典型应用
遍历输入字符串，遇到一个运算符就分别遍历运算符的左边和右边，通过这样的递归就能求到该运算符左边和右边的所有可能的值， 递归的终止情况是，字符串中只有数字；
1 class Solution { 2 public: 3 vector<int> diffWaysToCompute(string input) { 4 vector<int> ans; 5 for(int i=0; i<input.size(); i++){ 6 if(input[i]==‘-‘ || input[i]==‘+‘ || input[i]==‘*‘){ 7 vector<int> l=diffWaysToCompute(input.substr(0, i)); 8 vector<int> r=diffWaysToCompute(input.substr(i+1)); 9 for(int j=0; j<l.size(); j++){ 10 for(int k=0; k<r.size(); k++){ 11 if(input[i]==‘-‘) ans.push_back(l[j]-r[k]); 12 else if(input[i]==‘+‘) ans.push_back(l[j]+r[k]); 13 else if(input[i]==‘*‘) ans.push_back(l[j]*r[k]); 14 } 15 } 16 } 17 } 18 if(ans.empty()) ans.push_back(stoi(input)); 19 return ans; 20 } 21 };
leetcode 241. Different Ways to Add Parentheses
标签：input public tin operator nbsp str 情况 push span
原文地址：https://www.cnblogs.com/mr-stn/p/9218921.html
