Leetcode 399除数求值

题目定义:

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

方式一(图):

class Solution {
    private Map<String,Integer> variables = new HashMap<>();
    private Integer nvars = 0;
    public double[] calcEquation(List<List<String>> equations,
       double[] values, List<List<String>> queries) {
        
        List<Pair<Integer,Double>>[] edges = getEdges(equations,values);
        int queriesCount = queries.size();
        double[] ret = new double[queriesCount];
        for(int i = 0; i < queriesCount; i++){
            List<String> query = queries.get(i);
            double result = -1.0;
            if(variables.containsKey(query.get(0)) && variables.containsKey(query.get(1))){
                int ia = variables.get(query.get(0));
                int ib = variables.get(query.get(1));
                if(ia == ib)
                    result = 1.0;
                else{
                    Queue<Integer> points = new LinkedList<>();
                    points.offer(ia);
                    double[] ratios = new double[nvars];
                    Arrays.fill(ratios,-1.0);
                    ratios[ia] = 1.0;
                    while(!points.isEmpty() && ratios[ib] < 0){
                        int x = points.poll();
                        for(Pair pair : edges[x]){
                            int y = (int) pair.getKey();
                            double val = (double) pair.getValue();
                            if(ratios[y] < 0){
                                ratios[y] = ratios[x] * val;
                                points.offer(y);
                            }
                        }
                    }
                    result = ratios[ib];
                }
            }
            ret[i] = result;
        }
        return ret;
    }

    //存储图的节点和边权值
    private List<Pair<Integer,Double>>[] getEdges(List<List<String>> equations,
      double[] values){
        int n = equations.size();
        for(int i = 0; i < n; i++){
            variables.putIfAbsent(equations.get(i).get(0),nvars++);
            variables.putIfAbsent(equations.get(i).get(1),nvars++);
        }
        List<Pair<Integer,Double>>[] edges = new List[nvars];
        for(int i = 0 ;i < nvars; i++){
            edges[i] = new ArrayList<>();
        }
        for(int i = 0; i < n; i++){
            int va = variables.get(equations.get(i).get(0));
            int vb = variables.get(equations.get(i).get(1));
            edges[va].add(new Pair(vb,values[i]));
            edges[vb].add(new Pair(va,1.0 / values[i]));
        }
        return edges;
    }
}

方式二(Floyd算法):

class Solution {
    public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) {
        int nvars = 0;
        Map<String,Integer> variables = new HashMap<>();
        for(int i = 0; i < equations.size(); i++){
            if(!variables.containsKey(equations.get(i).get(0)))
                variables.put(equations.get(i).get(0),nvars++);
            if(!variables.containsKey(equations.get(i).get(1)))
                variables.put(equations.get(i).get(1),nvars++);
        }
        double[][] graph = new double[nvars][nvars];
        for(int i = 0 ; i < nvars; i++)
            Arrays.fill(graph[i],-1);
        for(int i = 0; i < equations.size(); i++){
            int va = variables.get(equations.get(i).get(0));
            int vb = variables.get(equations.get(i).get(1));
            graph[va][vb] = values[i];
            graph[vb][va] = 1.0 / values[i];
        }
        
        //floyd算法
        for(int k = 0; k < nvars; k++)
            for(int i = 0; i < nvars; i++)
                for(int j = 0; j < nvars; j++)
                    if(graph[i][k] > 0 && graph[k][j] > 0)
                        graph[i][j] = graph[i][k] * graph[k][j];
                        
        
        double[] ret  = new double[queries.size()];
        for(int i = 0; i < queries.size(); i++){
            List<String> query = queries.get(i);
            double result = -1.0;
            if(variables.containsKey(query.get(0)) && variables.containsKey(query.get(1))){
                int va = variables.get(query.get(0));
                int vb = variables.get(query.get(1));
                if(graph[va][vb] > 0)
                    result = graph[va][vb];
            }
            ret[i] = result;
        }
        return ret;
    }
}

方式三():


参考:

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