Divide and Conquer -- Leetcode problem241:Different Ways to Add Parentheses

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 *.

Example1

input:2-1-1

((2-1)-1) = 0

(2-(1-1)) = 2

output：[0,2]

Example2

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]

分析：这道题顾名思义是给表达式添加符号的问题，如果穷举出各种可能的加括号情况，很可能会产生重复或者遗漏的问题。并且这道题只需要加圆括号，因此加括号的位置不用考虑。所以可以把这个String类型的表达式根据每一运算符分为左右两边分别进行运算。

思路一：直接选用分治法进行运算。

class Solution {
public:
vector<int> diffWaysToCompute(string input) {
vector<int> result;
int size = input.size();
for (int i = 0; i < size; i++) {
char temp = input[i];
if (temp == '+' || temp == '-' || temp == '*') {
vector<int> left = diffWaysToCompute(input.substr(0, i));
vector<int> right = diffWaysToCompute(input.substr(i+1));
for (int j = 0 ; j < left.size(); j++) {
for (int k = 0; k < right.size(); k++) {
if (temp == '+')
result.push_back(left[j] + right[k]);
else if (temp == '-')
result.push_back(left[j] - right[k]);
else
result.push_ba
4000
ck(left[j] * right[k]);
}
}
}
}
if (result.empty())
result.push_back(stoi(input));
return result;
}
};