Leetcode-并查集/BFS/DFS-399. 除法求值

题目：

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

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/evaluate-division
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

题解：

一开始只想到DFS做法，把并查集全忘了，此处记录一下官方带权值的并查集做法

代码：

import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class Solution {

    public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) {
        int equationsSize = equations.size();

        UnionFind unionFind = new UnionFind(2 * equationsSize);
        // 第 1 步：预处理，将变量的值与 id 进行映射，使得并查集的底层使用数组实现，方便编码
        Map<String, Integer> hashMap = new HashMap<>(2 * equationsSize);
        int id = 0;
        for (int i = 0; i < equationsSize; i++) {
            List<String> equation = equations.get(i);
            String var1 = equation.get(0);
            String var2 = equation.get(1);

            if (!hashMap.containsKey(var1)) {
                hashMap.put(var1, id);
                id++;
            }
            if (!hashMap.containsKey(var2)) {
                hashMap.put(var2, id);
                id++;
            }
            unionFind.union(hashMap.get(var1), hashMap.get(var2), values[i]);
        }

        // 第 2 步：做查询
        int queriesSize = queries.size();
        double[] res = new double[queriesSize];
        for (int i = 0; i < queriesSize; i++) {
            String var1 = queries.get(i).get(0);
            String var2 = queries.get(i).get(1);

            Integer id1 = hashMap.get(var1);
            Integer id2 = hashMap.get(var2);

            if (id1 == null || id2 == null) {
                res[i] = -1.0d;
            } else {
                res[i] = unionFind.isConnected(id1, id2);
            }
        }
        return res;
    }

    private class UnionFind {

        private int[] parent;

        /**
         * 指向的父结点的权值
         */
        private double[] weight;


        public UnionFind(int n) {
            this.parent = new int[n];
            this.weight = new double[n];
            for (int i = 0; i < n; i++) {
                parent[i] = i;
                weight[i] = 1.0d;
            }
        }

        public void union(int x, int y, double value) {
            int rootX = find(x);
            int rootY = find(y);
            if (rootX == rootY) {
                return;
            }

            parent[rootX] = rootY;
          	// 关系式的推导请见「参考代码」下方的示意图
            weight[rootX] = weight[y] * value / weight[x];
        }

        /**
         * 路径压缩
         *
         * @param x
         * @return 根结点的 id
         */
        public int find(int x) {
            if (x != parent[x]) {
                int origin = parent[x];
                parent[x] = find(parent[x]);
                weight[x] *= weight[origin];
            }
            return parent[x];
        }

        public double isConnected(int x, int y) {
            int rootX = find(x);
            int rootY = find(y);
            if (rootX == rootY) {
                return weight[x] / weight[y];
            } else {
                return -1.0d;
            }
        }
    }
}

class Solution {
    public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) {
        int nvars = 0;
        Map<String, Integer> variables = new HashMap<String, Integer>();

        int n = equations.size();
        for (int i = 0; i < n; 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++);
            }
        }
        int[] f = new int[nvars];
        double[] w = new double[nvars];
        Arrays.fill(w, 1.0);
        for (int i = 0; i < nvars; i++) {
            f[i] = i;
        }

        for (int i = 0; i < n; i++) {
            int va = variables.get(equations.get(i).get(0)), vb = variables.get(equations.get(i).get(1));
            merge(f, w, va, vb, values[i]);
        }
        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)), ib = variables.get(query.get(1));
                int fa = findf(f, w, ia), fb = findf(f, w, ib);
                if (fa == fb) {
                    result = w[ia] / w[ib];
                }
            }
            ret[i] = result;
        }
        return ret;
    }

    public void merge(int[] f, double[] w, int x, int y, double val) {
        int fx = findf(f, w, x);
        int fy = findf(f, w, y);
        f[fx] = fy;
        w[fx] = val * w[y] / w[x];
    }

    public int findf(int[] f, double[] w, int x) {
        if (f[x] != x) {
            int father = findf(f, w, f[x]);
            w[x] = w[x] * w[f[x]];
            f[x] = father;
        }
        return f[x];
    }
}

 

 

说明：代码 weight[rootX] = weight[y] * value / weight[x]; 的推导过程，主要需要明白各个变量的含义，由两条路径有向边的权值乘积相等得到相等关系，然后做等价变换即可。

复杂度分析：

时间复杂度：O((N + Q)\log A)

构建并查集 O(NlogA) ，这里 N 为输入方程 equations 的长度，每一次执行合并操作的时间复杂度是 O(\log A) 这里 A是 equations 里不同字符的个数；
查询并查集 O(QlogA)，这里 Q 为查询数组 queries 的长度，每一次查询时执行「路径压缩」的时间复杂度是 O(\log A)
空间复杂度：O(A)：创建字符与 id 的对应关系 hashMap 长度为 A，并查集底层使用的两个数组 parent 和 weight 存储每个变量的连通分量信息，parent 和 weight 的长度均为 A。

 

 

BFS解法：

class Solution {
    public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) {
        int nvars = 0;
        Map<String, Integer> variables = new HashMap<String, Integer>();

        int n = equations.size();
        for (int i = 0; i < n; 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++);
            }
        }

        // 对于每个点，存储其直接连接到的所有点及对应的权值
        List<Pair>[] edges = new List[nvars];
        for (int i = 0; i < nvars; i++) {
            edges[i] = new ArrayList<Pair>();
        }
        for (int i = 0; i < n; i++) {
            int va = variables.get(equations.get(i).get(0)), 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]));
        }

        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)), ib = variables.get(query.get(1));
                if (ia == ib) {
                    result = 1.0;
                } else {
                    Queue<Integer> points = new LinkedList<Integer>();
                    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 = pair.index;
                            double val = pair.value;
                            if (ratios[y] < 0) {
                                ratios[y] = ratios[x] * val;
                                points.offer(y);
                            }
                        }
                    }
                    result = ratios[ib];
                }
            }
            ret[i] = result;
        }
        return ret;
    }
}

class Pair {
    int index;
    double value;

    Pair(int index, double value) {
        this.index = index;
        this.value = value;
    }
}

复杂度分析

时间复杂度：O(ML+Q\cdot(L+M))，其中 M为边的数量，Q 为询问的数量，L 为字符串的平均长度。构建图时，需要处理 M 条边，每条边都涉及到 O(L) 的字符串比较；处理查询时，每次查询首先要进行一次 O(L) 的比较，然后至多遍历 O(M) 条边。

空间复杂度：O(NL+M)，其中 N 为点的数量，M 为边的数量，L 为字符串的平均长度。为了将每个字符串映射到整数，需要开辟空间为 O(NL) 的哈希表；随后，需要花费 O(M) 的空间存储每条边的权重；处理查询时，还需要 O(N)的空间维护访问队列。最终，总的复杂度为 O(NL+M+N) = O(NL+M)

 

DFS

自己写的，跟BFS思路差不多，深搜

class Solution {
    public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries)    {
        
        HashMap<String, HashMap<String,Double>> map=new HashMap<String, HashMap<String,Double>>(); 
        double[] res=new double[queries.size()];
        int all=0;
        for( int i = 0 ; i < equations.size() ; i++) 
        {
            String a =  equations.get(i).get(0);
            String b =  equations.get(i).get(1);
            double v =  values[i];
            
            HashMap<String,Double> divisor1 = new HashMap<String,Double>();   
            divisor1.put(b,v);
            HashMap<String,Double> divisor2 = new HashMap<String,Double>();   
            divisor2.put(a,1/v);
            if(!map.containsKey(a))
            {
                map.put(a,divisor1);
                all+=1;
            }
                HashMap<String,Double> av = map.get(a);
                av.put(b,v);
                map.put(a,av);
            if(!map.containsKey(b))
            {
                map.put(b,divisor2);all+=1;
            }
                HashMap bv = map.get(b);
                bv.put(a,1/v);
                map.put(b,bv);
        }
         for( int i = 0 ; i < queries.size() ; i++) 
        {            
            String a =  queries.get(i).get(0);
            String b =  queries.get(i).get(1);
            if(map.containsKey(a)&&map.containsKey(b))
            {
                if(map.get(a).containsKey(b))
                {
                    res[i]=map.get(a).get(b);
                }
                if(a.equals(b))
                {
                    res[i]=1.0;
                }
                else
                {
                    Stack<String> st = new Stack<String>();
                    st.push(a);
                    HashMap<String,Double> vis = new HashMap<String,Double>(); 
                    for(String key : map.keySet())
                    {
                        vis.put(key,-1.0);
                    }
                    vis.put(a,1.0);
                    while (!st.isEmpty() && vis.get(b) < 0) {
                        String x = st.pop();
                        for(String key : map.get(x).keySet())
                        {
                            String y = key;
                            double val = map.get(x).get(y);
                            if (vis.get(y) < 0) 
                            {
                                vis.put(y,vis.get(x) * val);
                                st.push(y);
                            }
                        }
                    }
                    res[i]=vis.get(b);
                }

            }
            else
            {
                res[i]=-1.0;
            }
            
        }

    return res;
       
        
    }
}

Floyd 算法
对于查询数量很多的情形，如果为每次查询都独立搜索一次，则效率会变低。为此，我们不妨对图先做一定的预处理，随后就可以在较短的时间内回答每个查询。

在本题中，我们可以使用Floyd 算法，预先计算出任意两点之间的距离

class Solution {
    public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) {
        int nvars = 0;
        Map<String, Integer> variables = new HashMap<String, Integer>();

        int n = equations.size();
        for (int i = 0; i < n; 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], -1.0);
        }
        for (int i = 0; i < n; i++) {
            int va = variables.get(equations.get(i).get(0)), vb = variables.get(equations.get(i).get(1));
            graph[va][vb] = values[i];
            graph[vb][va] = 1.0 / values[i];
        }

        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];
                    }
                }
            }
        }

        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)), ib = variables.get(query.get(1));
                if (graph[ia][ib] > 0) {
                    result = graph[ia][ib];
                }
            }
            ret[i] = result;
        }
        return ret;
    }
}

 

 

 

本文地址：https://blog.csdn.net/qq_40421671/article/details/112274394