【Leetcode】241. Different Ways to Add Parentheses

题目：

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]

Credits:

Special thanks to @mithmatt for adding this problem and creating all test cases.

分析：

这道题的题意是给一个带有数字和运算符的字符串，得到该算式所有可能的答案。

我的思路是使用分治算法。将每个字符串分成两半再通过递归计算，将左右算式可能得到的结果保存在两个向量中，再通过两个for循环加在一起得到所有可能的答案。

代码：

class Solution {
public:
vector<int> diffWaysToCompute(string input) {
vector<int> l,r;
vector<int> result;      //保存算式可能的所有答案
result.clear();
int tt;
int tag = 0;   //标记字符串中是否存在运算符
for(int i = 0;i<input.length();i++)
{
if(input[i] == '+'||input[i] == '*'||input[i] == '-')     //将字符串分成两半通过递归计算
{
tag = 1;
string temp;
int j;
for(j = 0;j<i;j++)
temp += input[j];
l = diffWaysToCompute(temp);
temp.clear();
for(j = 0;j<input.length()-i-1;j++)
temp += input[j+i+1];
r = diffWaysToCompute(temp);
for(int p = 0;p<l.size();p++)
{
for(int q = 0;q<r.size();q++)
{
if(input[i] == '+')
{
tt = l[p]+r[q];
result.push_back(tt);
}
else if(input[i] == '-')
{
tt = l[p]-r[q];
result.push_back(tt);
}
else if(input[i] == '*')
{
tt = l[p]*r[q];
result.push_back(tt);
}
}
}

}

}
if(tag == 0)      //该字符串中不存在运算符号
{
int num = 0,temp;
int j = 0;
for(int i = input.length()-1;i>=0;i--)
{
temp = input[j]-'0';
num += temp*pow(10,i);
j++;
}
result.push_back(num);

}
return result;

}
};