Week Training: 241 Different Ways to Add Parentheses

最新推荐文章于 2020-08-10 02:49:26 发布

richard978

最新推荐文章于 2020-08-10 02:49:26 发布

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本文链接：https://blog.youkuaiyun.com/richard978/article/details/60466102

本文介绍了一种使用递归和分治策略解决复杂数学表达式计算问题的方法。通过将原始字符串分解为若干子字符串，并对每个子字符串递归应用相同的过程来计算可能的结果。文章详细解释了如何处理加法、减法和乘法运算符。

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As we are learning divide and conquer, a problem like this can help us understand well. It seems hard, since we don't know how long the string is, there might be lots of conditions. However, through the theory of divide and conquer and recursive, we can make the problem concise and easier to deal with. If we meet with operators like '+', '-' and '*', we just divide the string before and after it to 2 substrings, resursively doing the same thing to them. For each 2 substrings, we compute every element of it according to the operator. The principle of this method is easy-understanding, but of course too much loop used causes long run time, which is to be improved.

class Solution {
public:
    vector<int> diffWaysToCompute(string input) {
        vector<int> result;
        for(int i=0;i<input.size();i++){
            char sym = input[i];
            if(sym == '+'||sym == '-'||sym == '*'){
                vector<int> r1 = diffWaysToCompute(input.substr(0,i));
                vector<int> r2 = diffWaysToCompute(input.substr(i+1));
                for(int j=0;j<r1.size();j++){
                    for(int k=0;k<r2.size();k++){
                        if(sym=='+')
                            result.push_back(r1[j]+r2[k]);
                        else if(sym=='-')
                            result.push_back(r1[j]-r2[k]);
                        else if(sym=='*')
                            result.push_back(r1[j]*r2[k]);
                    }
                }
            }
        }
        if(result.empty())
            result.push_back(atoi(input.c_str()));
        return result;
    }
};