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]
这道题让给我们一个可能含有加减乘的表达式，让我们在任意位置添加括号，求出所有可能表达式的不同值。这道题跟之前的那道用的方法一样，用递归来解，划分左右子树，递归构造。
class Solution {public: vector diffWaysToCompute(string input) { vector res; for (int i = 0; i < input.size(); ++i) { if (input[i] == '+' || input[i] == '-' || input[i] == '*') { vector left = diffWaysToCompute(input.substr(0, i)); vector right = diffWaysToCompute(input.substr(i + 1)); for (int j = 0; j < left.size(); ++j) { for (int k = 0; k < right.size(); ++k) { if (input[i] == '+') res.push_back(left[j] + right[k]); else if (input[i] == '-') res.push_back(left[j] - right[k]); else res.push_back(left[j] * right[k]); } } } } if (res.empty()) res.push_back(atoi(input.c_str())); return res; }}; 本文转自博客园Grandyang的博客，原文链接：，如需转载请自行联系原博主。