leetcode:679. 24 Game

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博主提到在LeetCode（lc）上遇到hard类题目，面试方要求不许用double类型，博主询问该如何编写代码，还提到要搜索所有可能情况，疑似回溯算法。

mgj 看到在lc上是hard类，我就算找了点借口安慰自己，，，

然而，面那边要求不许用double我也是醉了。

怎么写？

jasonshieh

208

Last Edit: October 2, 2018 11:07 AM

2.0K VIEWS

Search for all possible cases.
Looks like backtracking...

class Solution {
public:
double elipson = pow(10.0, -5);
vector<char> operations = {'+','-','*','/'};
bool judgePoint24(vector<int>& nums) {
    vector<double> vec;
    for(auto n : nums){
        vec.push_back(n*1.0);
    }
    return find24(vec);
}

bool find24(vector<double> vec){
    if(vec.size() == 1){
        return abs(vec[0] - 24.0) <= elipson;
    }
    for(int i = 0; i < vec.size(); ++i){//each time we pick up two number for computation
        for(int j = 0; j < vec.size(); ++j){
            if(i == j) continue;
            vector<double> res;
            for(int k = 0; k < vec.size(); ++k){
                if(k != i && k != j) res.push_back(vec[k]);
            }
            for(auto op : operations){
                if((op == '+' || op == '*') && i > j) continue;//no need to re-calculate
                if(op =='/'  && vec[j] == 0.0) continue;
                switch(op){
                    case '+': res.push_back(vec[i] + vec[j]); break;
                    case '-': res.push_back(vec[i] - vec[j]); break;
                    case '*': res.push_back(vec[i] * vec[j]); break;
                    case '/': res.push_back(vec[i] / vec[j]); break;
                }
                if(find24(res)) return true;
                res.pop_back();//!!!important
            }
        }
    }
    return false;
}
};