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LeetCode-399. 除法求值

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题目来源

399. 除法求值

题目详情

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]values[i] 共同表示等式 Ai / Bi = values[i] 。每个 AiBi 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。如果问题中出现了给定的已知条件中没有出现的字符串,也需要用 -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:

题解分析

解法一:并查集

class Solution {

    public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) {
        HashMap<String, Integer> map = new HashMap<>();
        MergeFindSet mergeFindSet  = new MergeFindSet(equations.size() * 2);
        int cnt = 0;
        for(int i=0; i < equations.size(); i++){
            List<String> list = equations.get(i);
            String start = list.get(0);
            String end = list.get(1);

            if(!map.containsKey(start)){
                map.put(start, cnt);
                cnt++;
            }
            if(!map.containsKey(end)){
                map.put(end, cnt);
                cnt++;
            }

            mergeFindSet.merge(map.get(start), map.get(end), values[i]);
        }

        double[] res = new double[queries.size()];
        for(int i=0; i<queries.size(); i++){
            List<String> query = queries.get(i);
            String start = query.get(0);
            String end = query.get(1);

            if(map.containsKey(start) && map.containsKey(end)){
                res[i] = mergeFindSet.isConnect(map.get(start), map.get(end));
            }else{
                res[i] = -1.0;
            }
        }
        return res;
    }
    class MergeFindSet{
        int[] parent;
        double[] weight;

        MergeFindSet(int n){
            parent = new int[n];
            weight = new double[n];
            for(int i=0; i<n; i++){
                parent[i] = i;
                weight[i] = 1.0;
            }
        }

        int find(int x){
            if(x == parent[x]){
                return parent[x];
            }else{
                int tempparent = parent[x];
                parent[x] = find(parent[x]);// 递归找到父节点
                weight[x] = weight[x] * weight[tempparent]; // 路径压缩,重新计算权重
                return parent[x];
            }
        }

        void merge(int a, int b, double value){
            int pa = find(a);
            int pb = find(b);
            if(pa != pb){
                parent[pa] = pb;
                weight[pa] = weight[b] * value / weight[a];
            }
        }

        double isConnect(int a, int b){
            int pa = find(a);
            int pb = find(b);
            if(pa == pb){
                return weight[a] / weight[b];
            }else{
                return -1.0;
            }
        }
    }

}

标签:map,get,start,values,queries,求值,399,equations,LeetCode
来源: https://www.cnblogs.com/GarrettWale/p/15739785.html