399. Evaluate Division

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<double>.

equations = [ ["a", "b"], ["b", "c"] ],
values = [2.0, 3.0],
queries = [ ["a", "c"], ["b", "a"], ["a", "e"], ["a", "a"], ["x", "x"] ].

已知a/b, b/c ...的值，估计a/c , b/a...的值。

用图的方法解决，a/b的值就是a->b的路径的权重，那么要估计a->c的值，就是求a->c的路径。

解决：

1、为输入创建一个表table，存放权重，为了索引方便，为每个节点匹配一个索引值index。

2、对于每个要估计的式子 x1/x2：

2.1、如果x1->x2路径存在，直接输出。

2.2、如果求x1->x2不存在，找到一个未访问过的点xj，使得x1->xj路径存在，求xj->x2的路径。

那么findPath(x1, x2) = findPath(x1, xj) * findPath(xj, x2)；

2.3、更新table并输出。

 1 class Solution {
 2 public:
 3     double findPath(vector<vector<double>> &table, vector<int> &visited, int start, int target) {
 4         double res = -1.0;
 5         for (int i=0; i<table.size(); ++i) {
 6             if (table[start][target] > 0.0)
 7                 break;
 8             if (i == start)
 9                 continue;
10             if (table[start][i] > 0.0 && visited[i] == 0) {
11                 visited[i] = 1;
12                 res = table[start][i] * findPath(table, visited, i, target);
13                 visited[i] = 0;
14                 table[start][target] = res > 0.0? res: -1.0;
15             } 
16         }
17         return table[start][target];
18     }
19     vector<double> calcEquation(vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries) {
20         int len = equations.size();
21         unordered_map<string, int> index;
22         int i = 0;
23         for (auto &equ : equations) {
24             if (index.find(equ.first) == index.end())
25                 index[equ.first] = i++;
26             if (index.find(equ.second) == index.end())
27                 index[equ.second] = i++;
28         }
29         vector<vector<double>> table(i, vector<double>(i, -1.0));
30         while (--i >= 0) table[i][i] = 1.0;
31         for (i=0; i<len; ++i) {
32             pair<string, string> equ = equations[i];
33             table[index[equ.first]][index[equ.second]] = values[i];
34             table[index[equ.second]][index[equ.first]] = 1.0 / values[i];
35         }
36         vector<double> res;
37         for (auto &qur : queries) {
38             string start = qur.first;
39             string target = qur.second;
40             vector<int> visited(table.size());
41             if (index.find(start)==index.end() || index.find(target)==index.end())
42                 res.push_back(-1.0);
43             else {
44                 visited[i] = 1;
45                 res.push_back(findPath(table, visited, index[start], index[target]));
46                 visited[i] = 0;
47             }
48         }    
49         return res;
50     }
51 };

猜你喜欢