[leetcode 241]Different Ways to Add Parentheses

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]

给定一个字符串含有数字字符和+、-、*（没说有没有空格，就当有空格处理）
代码一多，就有点乱，思路大体是：
1、将输入字符串按照数字和运算符号分离，按顺序分别存入两个vector;
2、使用二维的vector作为标记，vector[i][j]表示字符串i到j的运算结果，因为运算结果可能有多个还有可能重复所以使用multiset
3、动态的计算vector[i][j]，知道得到结果vector[0][input.size()-1];
AC代码：

class Solution
{
public:
vector<int> diffWaysToCompute(string input)
{
int sum=0;
vector<int> num;
vector<char> sign;
int len=input.size();
int x=0;
int y=0;
vector<int> res;
while(x<len)
{
if(input[x]>='0'&&input[x]<='9')
{
y=0;
while(x<len&&input[x]>='0'&&input[x]<='9')
{
y=y*10+input[x]-48;
++x;
}
num.push_back(y);
++sum;
}
else if(input[x]==' ')
++x;
else
{
sign.push_back(input[x]);
++x;
}
}
vector<vector<multiset<int> > > temp;
multiset<int> temp_set;
vector<multiset<int> >temp_vec;
for(int i=0; i<sum; ++i)
temp_vec.push_back(temp_set);
for(int i=0; i<sum; ++i)
temp.push_back(temp_vec);
for(int i=0; i<sum; ++i)
temp[i][i].insert(num[i]);
for(int i=1; i<sum; ++i)
{
if(sign[i-1]=='+')
temp[i-1][i].insert(num[i-1]+num[i]);
else if(sign[i-1]=='-')
temp[i-1][i].insert(num[i-1]-num[i]);
else
temp[i-1][i].insert(num[i-1]*num[i]);
}
for(int i=2; i<sum; ++i)
{
for(int j=0; j+i<sum; ++j)
{
multiset<int>::iterator ite=temp[j][j].begin();
for(multiset<int>::iterator ite2=temp[j+1][i+j].begin(); ite2!=temp[j+1][i+j].end(); ++ite2)
{
if(sign[j]=='+')
temp[j][i+j].insert(*ite+(*ite2));
else if(sign[j]=='-')
temp[j][i+j].insert(*ite-(*ite2));
else
temp[j][i+j].insert(*ite*(*ite2));
}
ite=temp[i+j][i+j].begin();
for(multiset<int>::iterator ite2=temp[j][i+j-1].begin(); ite2!=temp[j][i+j-1].end(); ++ite2)
{
if(sign[i+j-1]=='+')
temp[j][i+j].insert(*ite2+(*ite));
else if(sign[j+i-1]=='-')
temp[j][i+j].insert(*ite2-(*ite));
else
temp[j][i+j].insert(*ite2*(*ite));
}
for(int k=j+1; k<i+j-1; ++k)
{
for(multiset<int>::iterator ite2=temp[j][k].begin(); ite2!=temp[j][k].end(); ++ite2)
{
for(multiset<int>::iterator ite3=temp[k+1][j+i].begin(); ite3!=temp[k+1][j+i].end(); ++ite3)
{
if(sign[k]=='+')
temp[j][i+j].insert(*ite2+(*ite3));
else if(sign[k]=='-')
temp[j][i+j].insert(*ite2-(*ite3));
else
temp[j][i+j].insert(*ite2*(*ite3));
}
}
}
}
}
for(multiset<int>::iterator ite2=temp[0][sum-1].begin();ite2!=temp[0][sum-1].end();++ite2)
res.push_back(*ite2);
return res;
}
};

其他Leetcode题目AC代码：https://github.com/PoughER/leetcode