iOS LeetCode ☞ 除法求值

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 <= 20equations[i].length == 21 <= Ai.length, Bi.length <= 5values.length == equations.length0.0 < values[i] <= 20.01 <= queries.length <= 20queries[i].length == 21 <= Cj.length, Dj.length <= 5Ai, Bi, Cj, Dj 由小写英文字母与数字组成 解题思路

没懂，抄的

代码

	// 399. 除法求值
    func calcEquation(_ equations: [[String]], _ values: [Double], _ queries: [[String]]) -> [Double] {
        var map = [String: [String: Double]]()
        var res = [Double]()
        for i in 0..<equations.count {
            let temp1 = equations[i][0]
            let temp2 = equations[i][1]
            let temp = values[i]
            var sub1 = map[temp1, default: [String: Double]()]
            sub1[temp2] = temp
            map[temp1] = sub1
            var sub2 = map[temp2, default: [String: Double]()]
            sub2[temp1] = 1 / temp
            map[temp2] = sub2
        }
        
        for q in queries {
            if let subMap1 = map[q[0]], let _ = map[q[1]] {
                var didChecks = [String]()
                var stack = [[String: Double]]()
                stack.append(subMap1)
                didChecks.append(q[0])
                var paths = [String]()
                paths.append(q[0])
                var isSuccess = false
                while stack.count > 0 {
                    let oMap = stack.last
                    var keyCount = 0
                    for key in oMap!.keys {
                        if key == q[1] {
                            // 可计算
                            paths.append(key)
                            stack.removeAll()
                            isSuccess = true
                            break
                        } else {
                            keyCount += 1
                            if !didChecks.contains(key) {
                                didChecks.append(key)
                                if let nextMap = map[key] {
                                    stack.append(nextMap)
                                    paths.append(key)
                                    break
                                } else {
                                    stack.removeLast()
                                    paths.removeLast()
                                }
                            } else {
                                if keyCount >= oMap!.keys.count {
                                    paths.removeLast()
                                    stack.removeLast()
                                }
                            }
                        }
                    }
                }
                if isSuccess {
                    var result = 1.0
                    for i in 0..<paths.count - 1 {
                        let tempMap = map[paths[i]]
                        let temp = tempMap![paths[i + 1]]
                        result *= temp!
                    }
                    res.append(result)
                } else {
                    res.append(-1.0)
                }
            } else {
                res.append(-1.0)
            }
        }
        return res
    }