leetcode241 : Different Ways to Add Parentheses

1、原题如下：

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]

2、解题如下：

class Solution {
public:
vector<int> diffWaysToCompute(string input) {
vector<int> result;
int offset=0;//用来弥补string中没有+-*的情况
for(int i=0;i<input.size();i++)
{
if(input[i]=='+'||input[i]=='-'||input[i]=='*')
{
offset=1;
vector<int> temp1=diffWaysToCompute(input.substr(0,i));
vector<int> temp2=diffWaysToCompute(input.substr(i+1));
for(auto x:temp1)
{
for(auto y:temp2)
{
if(input[i]=='+')
result.push_back(x+y);
if(input[i]=='-')
result.push_back(x-y);
if(input[i]=='*')
result.push_back(x*y);
}
}
}
}
if(offset==0)
{
result.push_back(stoi(input));
}
return result;
}
};

3、总结

循环迭代的算法题目很多，做到这会儿已经更加轻车熟路了，只需要注意边界条件即可，这里采用了offset来讨论边界问题