[leetcode]241. Different Ways to Add Parentheses

divide and conquer

题目描述：

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]

思路：

采用分而治之的方法，每遇到一个运算符，就将这个运算符左右两边的表达式看成两个子问题，递归调用函数得到左右两边表达式的运算结果集；然后根据该运算符对两个结果集进行交叉运算，放入最终的结果集。如果不是运算符，即结果集为0，则把字符串转化为int，放入结果集。

class Solution {
public:
vector<int> diffWaysToCompute(string input) {
vector<int> res;

int i;
for (i = 0; i < input.size(); i++) {
char c = input[i];
if (c == '-' || c == '*' || c == '+') {
vector<int> left = diffWaysToCompute(input.substr(0, i));
vector<int> right = diffWaysToCompute(input.substr(1+i));
for (int j = 0; j < left.size(); j++) {
for (int k = 0; k < right.size(); k++) {
if (c == '-') res.push_back(left[j]-right[k]);
if (c == '+') res.push_back(left[j]+right[k]);
if (c == '*') res.push_back(left[j]*right[k]);
}
}
}
}

if (res.size() == 0) res.push_back(atoi(input.c_str()));
return res;
}
};