399. Evaluate Division

399. Evaluate Divisionopen in new window

🟠 🔖 深度优先搜索 广度优先搜索 并查集 图 数组 最短路 🔗 LeetCodeopen in new window

题目

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.

Example 1:

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]
Explanation:
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 ]
note: x is undefined => -1.0

Example 2:

Input: equations = [["a","b"],["b","c"],["bc","cd"]], values = [1.5,2.5,5.0], queries = [["a","c"],["c","b"],["bc","cd"],["cd","bc"]]
Output: [3.75000,0.40000,5.00000,0.20000]

Example 3:

Input: equations = [["a","b"]], values = [0.5], queries = [["a","b"],["b","a"],["a","c"],["x","y"]]
Output: [0.50000,2.00000,-1.00000,-1.00000]

Constraints:

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 consist of lower case English letters and digits.

题目大意

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。

注意：未在等式列表中出现的变量是未定义的，因此无法确定它们的答案。

解题思路

这道题在考察 有向加权图 的遍历。

a/b=2 就相当于往图中添加了一条 a->b 权值为 2 的边，同时添加一条 b->a 权值为 1/2 的边。

queries 问 x/y 的值，相当于就是图中是否存在一条从 x 到 y 的路径，如果有，那么路径上所有边的权值相乘就是 x/y 的值。

所以思路就是，用邻接表建图，然后用 DFS 或者 BFS 遍历即可。

代码

/**
 * @param {string[][]} equations
 * @param {number[]} values
 * @param {string[][]} queries
 * @return {number[]}
 */
var calcEquation = function (equations, values, queries) {
	// 把 equations 抽象成一幅图，邻接表存储
	let graph = new Map();
	for (let i = 0; i < equations.length; i++) {
		let [start, end] = equations[i],
			weight = values[i];

		// 构建双向图
		if (!graph.has(start)) {
			graph.set(start, []);
		}
		graph.get(start).push({ node: end, weight });

		if (!graph.has(end)) {
			graph.set(end, []);
		}
		graph.get(end).push({ node: start, weight: 1 / weight });
	}
	// 依次求解 queries
	let res = new Array(queries.lenghth);
	for (let i = 0; i < queries.length; i++) {
		const [start, end] = queries[i];
		// BFS 遍历图，计算 start 到 end 的路径乘积
		res[i] = dfs(graph, start, end);
	}
	return res;
};

var dfs = function (graph, start, end) {
	// 不存在的节点，肯定无法到达
	if (!graph.has(start) || !graph.has(end)) return -1.0;
	if (start == end) return 1.0;

	// BFS 标准框架
	let queue = [start],
		visited = new Set([start]),
		// key 为节点变量名，value 记录从 start 到该节点的路径乘积
		weight = new Map([[start, 1.0]]);
	while (queue.length) {
		let cur = queue.shift();
		for (let item of graph.get(cur)) {
			if (visited.has(item.node)) {
				continue;
			}
			// 更新路径乘积
			weight.set(item.node, weight.get(cur) * item.weight);
			if (item.node == end) {
				return weight.get(end);
			}
			// 记录 visited
			visited.add(item.node);
			// 新节点加入队列继续遍历
			queue.push(item.node);
		}
	}
	return -1.0;
};

399. Evaluate Division

# 399. Evaluate Divisionopen in new window

# 题目

# 题目大意

# 解题思路

# 代码

# 相关题目

399. Evaluate Divisionopen in new window

题目

题目大意

解题思路

代码

相关题目