近期CSOahelp石老师的学生在tiktok的面试中，碰到一题典型的“除法方程”题——LeetCode 399。题目长这样（完整英文原文）：

Title: Division

Question description
You are given an array of variable pairs equations and an array of real numbers values, where
equations[i] = [Ai, Bi] and values[i] represent the equation Ai / Bi = values[i].
Each Ai or Bi is a string that represents a single variable.

You are also given some queries, where queries[j] = [Cj, Dj] represents the jth query
where you must find the answer for Cj / Dj = ?. Return the answers to all queries.
If a single answer cannot be determined, return -1.0.

Note: The input is always valid. You may assume that evaluating the queries will not result
in division by zero and that there is no contradiction.
Note: The variables that do not occur in the list of equations are undefined,
so the answer cannot be determined for them.

Input:
equations = [["a","b"],["b","c"]],
values = [2.0,3.0],
queries = [["a","c"],["b","a"],["a","e"],["a","a"],["x","x"]]

Output: [6.00000,0.50000,-1.00000,1.00000,-1.00000]

一句话拆解全题：

把变量当节点，除法当有向带权边，用 DFS 找路径、连乘权重。

下面是一个伪代码的示例，

        
from collections import defaultdict

def calc(eqs, vals, qs):
    # 1. 建图：g[X][Y] = X/Y；反向 g[Y][X] = 1/v
    g = defaultdict(dict)
    for (X, Y), v in zip(eqs, vals):
        g[X][Y] = v
        g[Y][X] = 1/v

    # 2. DFS：从 u 到 t，沿途乘权重，找不到就 -1
    def dfs(u, t, seen):
        if u not in g or t not in g:       # 变量没出现过
            return -1.0
        if u == t:                         # x/x = 1
            return 1.0
        seen.add(u)
        for w, wt in g[u].items():
            if w in seen: continue
            res = dfs(w, t, seen)
            if res != -1.0:
                return wt * res
        return -1.0

    # 3. 一行搞定所有查询
    return [dfs(a, b, set()) for a, b in qs]

# 测试一下
eqs = [["a","b"],["b","c"]]
vals= [2.0,3.0]
qs  = [["a","c"],["b","a"],["a","e"],["a","a"],["x","x"]]
print(calc(eqs, vals, qs))  # [6.0, 0.5, -1.0, 1.0, -1.0]
        
    

核心要点

• 建图：g[X][Y] = v，以及 g[Y][X] = 1/v
• DFS：带一个 seen 集合防环，找到目标就把沿路所有 wt 相乘
• 边界：变量不存在立刻 -1.0；自除 x/x = 1.0

如果想再快一点，可以加个小缓存：DFS 找到一次结果后，直接 g[C][D] = 结果，下一次同样查询瞬间返回。或者用并查集（Union-Find）带权优化，查询能做到几乎 O(1)。

题解leetcode上有很多写的比我好的，我就不献丑讲更多了。

2025暑期发面试的公司不多，tiktok算是其中比较活跃的了，如果你暑期收到了tiktok的面试，不妨认真准备一下，竞争对手其实会比秋招的时候少很多。

另外暑期活跃的公司还有Amazon，需要Amazon亚麻真题的可以看这里：传送门

如果同学们对于在北美求职有什么疑问，欢迎咨询我哦，石老师知无不言。24小时欢迎大家来骚扰。

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