241. Different Ways to Add Parentheses 分治法的应用

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.

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分析：

对于每个符号做分治法来考虑。对于每个符号做分治法，运用递归的方法来解决问题。此题解法是抄自别人的。只能验证他的正确性。

代码：

class Solution {
public:
    vector<int> diffWaysToCompute(string input) {
        vector<int> v;
        for(int i=0;i<input.size();++i)
        {
            char p=input[i];
            if(p=='+'||p=='-'||p=='*')
            {
                vector<int> v1=diffWaysToCompute(input.substr(0,i));
                vector<int> v2=diffWaysToCompute(input.substr(i+1));
                for(auto p1:v1)
                for(auto p2:v2)
                {
                    if(p=='+')
                    v.push_back(p1+p2);
                    if(p=='-')
                    v.push_back(p1-p2);
                    if(p=='*')
                    v.push_back(p1*p2);
                }
            }
        }
        if(v.empty())
        {
            int sum=0;
            for(int i=0;i<input.size();++i)
            {
                sum=sum*10+input[i]-'0';
            }
            v.push_back(sum);
        }
        return v;
    }
};