LeetCode 399. 除法求值
作者:互联网
题目描述
给出方程式 A / B = k
, 其中 A
和 B
均为用字符串表示的变量, k
是一个浮点型数字。根据已知方程式求解问题,并返回计算结果。如果结果不存在,则返回 -1.0
。
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:
给定:a / b = 2.0, b / c = 3.0
问题:a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ?
返回:[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 <= equations[i][0].length, equations[i][1].length <= 5
values.length == equations.length
0.0 < values[i] <= 20.0
1 <= queries.length <= 20
queries[i].length == 2
1 <= queries[i][0].length, queries[i][1].length <= 5
equations[i][0], equations[i][1], queries[i][0], queries[i][1]
由小写英文字母与数字组成
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/evaluate-division
思路解析
深度优先搜索(DFS)
unordered_map<string, unordered_map<string, double>> graph
表示图,可以通过访问graph['a']['b']
得到\(a/b\)的结果(如果存在)。unordered_set<string> visited
表示已经访问过的节点。- 若要计算\(a / b\):
- 查找是否存在
graph['a']['b']
,若存在,直接返回结果; - 若不存在
graph['a']['b']
,则查找graph['a']
中存在哪些节点,比如存在'c'
,'d'
...,则分别查找graph['c']['b']
,graph['d']['b']
- 倘若从
graph['c']['b']
中找到了结果,记为t
,则返回t * graph['a']['c']
- 若都没有找到结果,返回
-1
。
- 查找是否存在
代码实现
class Solution {
private:
unordered_map<string, unordered_map<string, double>> graph;
double dfs(string src, string dst, unordered_set<string> visited) {
if(graph[src].count(dst))
return graph[src][dst];
for(const auto subequation : graph[src]) {
if(visited.count(subequation.first))
continue;
visited.insert(subequation.first);
double t = dfs(subequation.first, dst, visited);
if(t > 0)
return t * subequation.second;
}
return -1;
}
public:
vector<double> calcEquation(vector<vector<string>>& equations, vector<double>& values, vector<vector<string>>& queries) {
for(int i = 0; i < equations.size(); i++) {
auto e = equations[i];
graph[e[0]][e[1]] = values[i];
graph[e[1]][e[0]] = 1.0 / values[i];
}
vector<double> result;
for(auto q : queries) {
unordered_set<string> visited;
double res = dfs(q[0], q[1], visited);
result.push_back(res);
}
return result;
}
};
标签:graph,1.00000,values,queries,求值,visited,399,equations,LeetCode 来源: https://www.cnblogs.com/xqmeng/p/13813795.html