目录

一、题目内容

二、解题思路

三、代码

一、题目内容

给你一个变量对数组 equations 和一个实数值数组 values 作为已知条件，其中 equations[i] = [Ai, Bi] 和 values[i] 共同表示等式 Ai / Bi = values[i] 。每个 Ai 或 Bi 是一个表示单个变量的字符串。

另有一些以数组 queries 表示的问题，其中 queries[j] = [Cj, Dj] 表示第 j 个问题，请你根据已知条件找出 Cj / Dj = ? 的结果作为答案。

返回 所有问题的答案 。如果存在某个无法确定的答案，则用 -1.0 替代这个答案。

注意：输入总是有效的。你可以假设除法运算中不会出现除数为 0 的情况，且不存在任何矛盾的结果。

示例 1：

输入：equations = [["a","b"],["b","c"]], values = [2.0,3.0], queries = [["a","c"],["b","a"],["a","e"],["a","a"],["x","x"]]
输出：[6.00000,0.50000,-1.00000,1.00000,-1.00000]
解释：
条件：a / b = 2.0, b / c = 3.0
问题：a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ?
结果：[6.0, 0.5, -1.0, 1.0, -1.0 ]

示例 2：

输入：equations = [["a","b"],["b","c"],["bc","cd"]], values = [1.5,2.5,5.0], queries = [["a","c"],["c","b"],["bc","cd"],["cd","bc"]]
输出：[3.75000,0.40000,5.00000,0.20000]

示例 3：

输入：equations = [["a","b"]], values = [0.5], queries = [["a","b"],["b","a"],["a","c"],["x","y"]]
输出：[0.50000,2.00000,-1.00000,-1.00000]

提示：

1 <= equations.length <= 20
equations[i].length == 2
1 <= Ai.length, Bi.length <= 5
values.length == equations.length
0.0 < values[i] <= 20.0
1 <= queries.length <= 20
queries[i].length == 2
1 <= Cj.length, Dj.length <= 5
Ai, Bi, Cj, Dj 由小写英文字母与数字组成

二、解题思路

看着很唬人，实际上建图即可，graph里增加的是每个分式的分子和分母的对应关系，weight里是分式的值，还有对应的倒数， visited存储遍历过的起始节点；

DFS，查找graph里的已有分式条件，以及利用中间节点构建新的分式，无法构建则为0，查询则返回-1.0；

三、代码

  

   

    

     

    
    

     

      from collections import defaultdict
     
    
   

    

     

    
    

     
 
     
    
   

    

     

    
    

     
 
     
    
   

    

     

    
    

     

      class Solution:
     
    
   

    

     

    
    

     
 def calcEquation(self, equations: list, values: list, queries: list) -> list:
     
    
   

    

     

    
    

     
 """
     
    
   

    

     

    
    

     

       :type equations: List[List[str]]
     
    
   

    

     

    
    

     

       :type values: List[float]
     
    
   

    

     

    
    

     

       :type queries: List[List[str]]
     
    
   

    

     

    
    

     

       :rtype: List[float]
     
    
   

    

     

    
    

     

       """
     
    
   

    

     

    
    

     

       gragh = defaultdict(set)
     
    
   

    

     

    
    

     

       weight = defaultdict()
     
    
   

    

     

    
    

     
 
     
    
   

    

     

    
    

     
 for i in range(len(equations)):
     
    
   

    

     

    
    

     

       gragh[equations[i][0]].add(equations[i][1])
     
    
   

    

     

    
    

     

       gragh[equations[i][1]].add(equations[i][0])
     
    
   

    

     

    
    

     

       weight[(equations[i][0], equations[i][1])] = values[i]
     
    
   

    

     

    
    

     

       weight[(equations[i][1], equations[i][0])] = float(1 / values[i])
     
    
   

    

     

    
    

     

       print(gragh)
     
    
   

    

     

    
    

     

       print(weight)
     
    
   

    

     

    
    

     
 
     
    
   

    

     

    
    

     
 def dfs(st, ed, visited):
     
    
   

    

     

    
    

     
 if (st, ed) in weight:
     
    
   

    

     

    
    

     
 return weight[(st, ed)]
     
    
   

    

     

    
    

     
 if st not in gragh or ed not in gragh or st in visited:
     
    
   

    

     

    
    

     
 return 0
     
    
   

    

     

    
    

     

       visited.add(st)
     
    
   

    

     

    
    

     
 
     
    
   

    

     

    
    

     

       res = 0
     
    
   

    

     

    
    

     
 for link in gragh[st]:
     
    
   

    

     

    
    

     

       res = (dfs(link, ed, visited)) * weight[(st, link)]
     
    
   

    

     

    
    

     
 if res != 0:
     
    
   

    

     

    
    

     

       weight[(st, ed)] = res
     
    
   

    

     

    
    

     
 break
     
    
   

    

     

    
    

     

       visited.remove(st)
     
    
   

    

     

    
    

     
 return res
     
    
   

    

     

    
    

     
 
     
    
   

    

     

    
    

     

       res = []
     
    
   

    

     

    
    

     
 for q in queries:
     
    
   

    

     

    
    

     

       ans = dfs(q[0], q[1], set())
     
    
   

    

     

    
    

     
 if ans == 0:
     
    
   

    

     

    
    

     

       ans = -1.0
     
    
   

    

     

    
    

     

       res.append(ans)
     
    
   

    

     

    
    

     
 return res
     
    
   

    

     

    
    

     
 
     
    
   

    

     

    
    

     
 
     
    
   

    

     

    
    

     

      if __name__ == '__main__':
     
    
   

    

     

    
    

     

       equations = [["a", "b"],
     
    
   

    

     

    
    

     

       ["b", "c"]]
     
    
   

    

     

    
    

     

       values = [2.0, 3.0]
     
    
   

    

     

    
    

     

       queries = [["a", "c"],
     
    
   

    

     

    
    

     

       ["b", "a"],
     
    
   

    

     

    
    

     

       ["a", "e"],
     
    
   

    

     

    
    

     

       ["a", "a"],
     
    
   

    

     

    
    

     

       ["x", "x"]]
     
    
   

    

     

    
    

     

       s = Solution()
     
    
   

    

     

    
    

     

       ans = s.calcEquation(equations, values, queries)
     
    
   

    

     

    
    

     

       print(ans)
     
    
  
 

文章来源: nickhuang1996.blog.csdn.net，作者：悲恋花丶无心之人，版权归原作者所有，如需转载，请联系作者。

原文链接：nickhuang1996.blog.csdn.net/article/details/112260851