399. [✔][M]除法求值

'[["a","c"],["b","a"],["a","e"],["a","a"],["x","x"]]'

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。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：

• 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

---

题解：

不太懂！！！

// @lc code=start
function calcEquation(equations: string[][], values: number[], queries: string[][]): number[] {
    class UnionFind {
        parent: Uint8Array
        weight: Float32Array
        constructor(n) {
            this.parent = new Uint8Array(n)
            this.weight = new Float32Array(n)
            while (n--) {
                this.parent[n] = n
                this.weight[n] = 1.0
            }
        }
        union(x, y, value) {
            const rootX = this.find(x), rootY = this.find(y)
            if (rootX !== rootY) {
                this.parent[rootX] = rootY
                this.weight[rootX] = this.weight[y] * value / this.weight[x]
            }
        }
        find(x) {
            if (x !== this.parent[x]) {
                const orginX = this.parent[x]
                this.parent[x] = this.find(this.parent[x])
                this.weight[x] *= this.weight[orginX]
            }
            return this.parent[x]
        }
        isConnected(x, y) {
            const rootX = this.find(x), rootY = this.find(y)
            return rootX !== void 0 && rootX === rootY ? this.weight[x] / this.weight[y] : -1.0
        }
    }

    const unionFind = new UnionFind(values.length << 1), h = new Map
    for (let i = 0, id = 0; i < values.length; i++) {
        const x = equations[i][0], y = equations[i][1]
        if (!Array.from(h.keys()).includes(x)) h.set(x, id++)
        if (!Array.from(h.keys()).includes(y)) h.set(y, id++)
        unionFind.union(h.get(x), h.get(y), values[i])
    }
    return queries.map(([x, y]) => unionFind.isConnected(h.get(x), h.get(y)))

};
// @lc code=end