问题地址：https://leetcode.com/problems/evaluate-division/description/

题意：

给2个关系数组，给出一系列除法的结果，如果没有结果，返回-1 。

Given：a / b = 2.0, b / c = 3.0.

queries are:a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ? .

return[6.0, 0.5, -1.0, 1.0, -1.0 ].

思路：

创建关系网，

a / b = 2.0 => a = b * 2.0 | b = a * 0.5 ，以此类推。

除此之外，还需要嵌套扩增关系网，

如果已有关系网中，a与b有关系，b与c有关系，c与d有关系，那关系网中必然还应该有：[a,c], [a,d], [b,d] 以上几个嵌套关系，具体代码如下：

/** * @param {string[][]} equations * @param {number[]} values * @param {string[][]} queries * @return {number[]} */var calcEquation = function(equations, values, queries){    var results = [];    var relations = new Map();        function arrangeRelations(){        for (var i = 0; i < values.length; ++i){            if (equations[i][0] === equations[i][1])                continue;            var keyword = equations[i][0];            if (!relations.has(keyword)){                relations.set(keyword, []);            }            relations.get(keyword).push([values[i], equations[i][1]]);                        keyword = equations[i][1];            if (!relations.has(keyword)){                relations.set(keyword, []);            }            relations.get(keyword).push([1 / values[i], equations[i][0]]);        }        enhanceRelations();    }    arrangeRelations();        function enhanceRelations() {        var addList = new Map();        relations.forEach((val1, key1) => {            for(let k = 0; k < val1.length; ++k) {                let key = val1[k][1];                if (relations.has(key)) {                    for (let p = 0; p < relations.get(key).length; ++p) {                        if (relations.get(key)[p][1] !== key1) {                            if (!addList.has(key1)) {                                addList.set(key1, new Set());                            }                            let vv = val1[k][0] * relations.get(key)[p][0];                            let nn = relations.get(key)[p][1];                            let contains = false;                            for (let x of addList.get(key1)) {                                if (x[1] === nn) {                                    contains = true;                                    break;                                }                            }                            for (let x of relations.get(key1)) {                                if (x[1] === nn) {                                    contains = true;                                    break;                                }                            }                            if (!contains)                                addList.get(key1).add([vv, nn]);                        }                    }                }            }        });        let length = 0;        addList.forEach((val, key) => {            if (!relations.has(key)) {                relations.set(key, [])            }            if (val.size) {                relations.set(key, relations.get(key).concat(Array.from(val)));                length ++;            }        });        debugger;        if (length) {            enhanceRelations();        }    }        for (var i = 0; i < queries.length; ++i) {        if (!relations.has(queries[i][0])) {            results.push(-1);            continue;        }        if (queries[i][0] === queries[i][1]) {            results.push(1);            continue;        }        var flag = false;        relations.forEach((val, key) => {            if (key === queries[i][0]) {                for(var k = 0;k < relations.get(key).length; ++k) {                    var tmp = relations.get(key)[k];                    if (tmp[1] === queries[i][1]) {                        flag = true;                        results.push(tmp[0]);                    }                }            }         });        if (!flag)            results.push(-1);    }    return results;};calcEquation([["x1","x2"],["x2","x3"],["x3","x4"],["x4","x5"]],[3.0,4.0,5.0,6.0],[["x1","x5"],["x5","x2"],["x2","x4"],["x2","x2"],["x2","x9"],["x9","x9"]])复制代码

Acceptd: