leetcode 399. Evaluate Division 等式求解+典型的DFS深度优先遍历

Equations are given in the format A / B = k, where A and B are variables represented as strings, and k is a real number (floating point number). Given some queries, return the answers. If the answer does not exist, return -1.0.

Example:

Given a / b = 2.0, b / c = 3.0.

queries are: a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ? .

return [6.0, 0.5, -1.0, 1.0, -1.0 ].

The input is: vector （pair（string, string）） equations, vector（double）& values, vector（pair（string, string）） queries , where equations.size() == values.size(), and the values are positive. This represents the equations. Return vector.

According to the example above:

equations = [ [“a”, “b”], [“b”, “c”] ],

values = [2.0, 3.0],

queries = [ [“a”, “c”], [“b”, “a”], [“a”, “e”], [“a”, “a”], [“x”, “x”] ].

The input is always valid. You may assume that evaluating the queries will result in no division by zero and there is no contradiction.

题意很简单，就是做一个除法的计算，就是一个DFS深度优先遍历的实现

这道题的要点是可以使用后继结点来保存树结构，使用父亲节点也可以，别的都没有难点了

注意这里的set是为了防止陷入死循环

代码如下：

#include <iostream>
#include <vector>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <numeric>
#include <cmath>
#include <regex>

using namespace std;

class Solution
{
public:
map<string, map<string, double>> m;
vector<double> calcEquation(vector<pair<string, string>> equations,
vector<double>& values, vector<pair<string, string>> query)
{
vector<double> res;
for (int i = 0; i < values.size(); ++i)
{
m[equations[i].first].insert(make_pair(equations[i].second, values[i]));
if (values[i] != 0)
m[equations[i].second].insert(make_pair(equations[i].first, 1 / values[i]));
}

for (auto i : query)
{
set<string> s;
double tmp = check(i.first, i.second, s);
res.push_back(tmp);
}
return res;
}

double check(string up, string down, set<string> &s)
{
if (m[up].find(down) != m[up].end())
return m[up][down];
else
{
for (auto i : m[up])
{
if (s.find(i.first) == s.end())
{
s.insert(i.first);
double tmp = check(i.first, down, s);
if (tmp != -1)
return i.second*tmp;
}
}
return -1;
}
}
};